Answers
Decrease.
What is the weight on the equator than at the Poles?Yes, You weigh less on the equator than at the North or South Pole, but the difference is small.
Note that your body itself does not change.
Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
In short, a trip to the equator is not a viable long-term weight-loss program.
Your weight is the combination of all the large-scale, long-term forces on your body. While the earth's gravity is by far the strongest large-scale force, it is not the only one.
We experience as "something pulling you down" is actually the total of all the forces and not just gravity.
The four dominant large-scale, long-term forces are:
The earth's gravity
The sun's gravity
The moon's gravity
The earth's centrifugal force
Hence, the weight of a body at the equator will be less than at the Poles.
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Related Questions
The unknown quantities to be determined are (a) the capacitive reactance, (b) the maximum and the rms voltages from the source, and (c) the rms current in the capacitor.
Evaluate the capacitive or inductive reactance, XC or XL.
The unknown quantity to be determined in part (a) is the circuit's capacitive reactance
XC defined as
XC ≡ 1/2fC
,
where, in this problem, C = 4.48 F and the AC source frequency f must be determined.
What is the frequency f of the AC source?
Answers
The frequency f of the AC source is determined as 0.446 Xc.
Frequency of the AC sourceThe frequency of the AC source is calculated as follows;
Xc ≡ 1/2fC
where;
Xc is the capacitive reactancef is frequencyC is capacitancefC = 2Xc
f = 2Xc / C
f = (2Xc)/4.48
f = 0.446 Xc
Thus, the frequency f of the AC source is determined as 0.446 Xc.
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Which job falls within the field of organizational psychology?
Job Recruiter
Employee Conflict Mediators
Psychiatrist
Receptionist
Answers
A car moves with an average speed of 75 kmh^-1 from town P to town Q in 2 hours. By using information, you may calculate the distance between two towns. state a derived quantity and its S.I unit
Answers
Answer:
patron
Explanation:
The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.
The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.
Explanation:
patron
Explanation:
The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.
The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.
D = (75)(2)
D = 150 km.
Therefore, the distance between town p and town q is 150 km.
A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hin ge. The rod can rotate around the hin ge without friction.
Initially the rod is held at rest at an angle of θ = 70.4° with respect to the horizontal surface. Then the rod is released.
1. What is the angular speed of the rod, when it lands on the horizontal surface?
2. What is the angular acceleration of the rod, just before it touches the horizontal surface?
Answers
Hello!
1.
To solve, we can use the work-energy theorem.
Initially, the rod is at rest. Its center of mass is some height above the surface, so it contains gravitational potential energy.
Finally, the rod has rotational kinetic energy.
To find the initial energy, we have to find where the center of mass is. It is located at the midpoint of the rod, or l/2. To find its height above, though, we have to find its sine component with respect to the angle:
[tex]E_i = mg{\frac{l}{2}}}sin(\theta)[/tex]
Its final energy:
[tex]E_f = \frac{1}{2}I\omega^2[/tex]
About the pivot point, a rod's moment of inertia (I) is 1/3ml².
[tex]E_f = \frac{1}{2}(\frac{1}{3}ml^2) = \frac{1}{6}ml^2\omega^2[/tex]
Now, solving for angular speed:
[tex]E_i = E_f\\\\mg\frac{l}{2}sin\theta = \frac{1}{6}ml^2 \omega^2\\\\3gsin\theta = l\omega^2 \\\\\omega = \sqrt{\frac{3gsin\theta}{l}}[/tex]
Plugging in the values and solving:
[tex]\omega = \sqrt{\frac{3(9.8)sin(70.4)}{2.1}} = \boxed{3.63 rad/s}[/tex]
2.
Let's determine where to place our pivot point. To minimize the number of torques to take into account, let's put the pivot point at the base of the rod.
Now, the only torque we must consider is that produced by the force of gravity.
Recall the equation for torque:
[tex]\tau = r \times F[/tex]
This is a cross-product, where 'r' is the lever arm of the force. This is the perpendicular distance between the line of action of the force and the pivot point.
In this instance, based on our reference angle, this would be the COSINE.
[tex]\tau = rFcos\theta[/tex]
r = distance from pivot point to center of the rod (1.05 m)
F = Force of gravity, mg (19.03 N)
θ = angle of rod from horizontal
Using the rotational equivalent of Newton's Second Law:
[tex]I\alpha = rFcos\theta[/tex]
'I' is the rod's moment of inertia about the pivot point. For a uniform rod, its moment of inertia at its end is 1/3ml².
Therefore:
[tex]I = \frac{1}{3}(1.94)(2.1^2) = 2.85 kgm^2[/tex]
Now, we can solve for the function for angular acceleration.
[tex]\alpha(\theta) = \frac{rF}{I}cos\theta\\\\\alpha(\theta) = \frac{(1.05)(19.03)}{2.85}cos\theta = 7cos\theta[/tex]
This is the angular acceleration of the rod as a function of theta.
Substituting in θ = 0°:
[tex]\alpha(0) = 7cos(0) = \boxed{7 rad/s^2}[/tex]
**Note, we couldn't just use rotational kinematics because the angular acceleration is NOT constant.
A person walks first at a constant speed of 5.00 m.s-1
along a straight line from point A to
point B and then back along the line from B to A at a constant speed of 3.00 m.s-1
.
(a) Calculate her average speed over the entire trip.
Answers
The average speed of the person over the entire trip is 3.75 m/s.
Average speed of the personAverage speed = total distance/total time
Average speed = (d + d)/(d/Va + d/Vb)
where;
d is the distance in a single tripVa is speed from point AVb is the speed from point BAverage speed = (2Vavb)/(Va + Vb)
Average speed = (2 x 5 x 3)/(5 + 3)
Average speed = 3.75 m/s
Thus, the average speed of the person over the entire trip is 3.75 m/s.
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Drag each label to the correct location on the table sort the sentences based on weather thy are correct or incorrect
Answers
The statement that water boils at a temperature below 100 degrees in a pressure cooker is correct
What is freezing?The term freezing refers to the temperature at which a liquid is turned into a solid.
The following is the sorting of the statements into those that are correct and incorrect.
Water boils at a temperature below 100 degrees at high altitudes - correct
Water expands on freezing - correct
Water contracts on freezing - incorrect
Water boils at a temperature below 100 degrees in a pressure cooker - correct
Water boils at 100 degrees under normal pressure - correct
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Two balls of equal mass collide and stick together as shown in the figure. The initial velocity of ball B is twice that of ball A. a. Calculate the angle above the horizontal of the motion of mass A + B after the collision. b. What is the ratio of the final velocity of the mass A + B to the initial velocity of ball A, /? c. What is the ratio of the final energy of the system to the initial energy of the system, /? Is the collision elastic or inelastic?
Answers
The angle above the horizontal of the motion of mass A + B after the collision. as well as others are mathematically given as
[tex]\theta f = 26.3 \deg[/tex]vf/vA= 0.942vf^2 / 5vA^2 = 0.35What is Collison equilibrium?
According to the collision hypothesis of chemistry, chemical reactions happen when two molecules or atoms collide.
Generally, the equation for Collison equilibrium is mathematically given as
For horizontal motion
mvAx+m2vAx = 2mvfx
For vertical motion
-mvAy+m2vAy = 2mvfy
vfy = 1/2 vAy
[tex]\tan \theta f = vfy/vfx \\\\\tan \theta f = 1/3 vAy/vAx \\\\\tan \theta f = vAy/vAx \\\\\theta f = tan-1 ( 1/3 tan (56) )\\\\[/tex]
[tex]\theta f = 26.3 \deg[/tex]
(b)
[tex]v fy = vf sin \theta f\\\\vAy = vA sin theta i\\\\vf/vA = vfy/vAy sin( \theta i)/sin(\theta f)\\\\vf/vA = 1/2 sin(56)/sin(26.3) \\\\[/tex]
vf/vA= 0.94
(c)
[tex]Ef/Ei = \frac{(1/2 * (2m) * (vf)^2) }{ (1/2 * m * (vA)^2) + 1/2*m*(2vA)^2) ]}[/tex]
2vf^2 / 5vA^2 = 0.35
In conclusion, The angle above the horizontal of the motion of mass A + B after the collision. as well as others
[tex]\theta f = 26.3 \deg[/tex]
vf/vA= 0.94
2vf^2 / 5vA^2 = 0.35
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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)
The ball is kicked with a speed of v0 = 15.10 m/s at an angle of θ = 74.1° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. When the ball lands on the other building, its speed is 19.89 m/s.
How much energy was lost to air friction? The ball is kicked without a spin.
37.71 J is incorrect.
Answers
The energy was lost to air friction is determined as 42.56 J.
Energy lost due to friction
The energy lost due to friction is calculated as follows;
ΔE = ΔK.E + ΔP.E
where;
ΔK.E is change in kinetic energyΔP.E is change in potential energyΔE = ¹/₂(0.45)(19.89² - 15.1²) + (0.45)(9.8)(12 - 30.2)
ΔE = 37.7 - 80.26
ΔE = -42.56 J
Thus, the energy was lost to air friction is determined as 42.56 J.
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a 600 g pool ball moving at 2.0m/s hits a second pool of the same mass at rest. the first ball moves off at 1.5m/s at an angle of 30 degrees from the original direction. assume the surface of the pool table to b frictionless find the angle at which the second ball moves off and the speed of the second ball after impact.
Answers
The speed of the second ball after impact is 1.43 m/s at 60.8⁰.
Speed of the second ballApply the principle of conservation of linear momentum;
in y direction0.6(2) = 0.6(1.5 sin30) + 0.6(vy)
1.2 = 0.45 + 0.6vy
0.6vy = 0.75
vy = 0.75/0.6
vy = 1.25 m/s
in x direction0.6(2) = 0.6(1.5 cos30) + 0.6(vx)
1.2 = 0.78 + 0.6vx
0.6vx = 0.42
vx = 0.42/0.6
vx = 0.7 m/s
Resultant speedv = √(vy² + vx²)
v = √(1.25² + 0.7²)
v = 1.43 m/s
Direction of second massθ = arc tan(vy/vx)
θ = arc tan (1.25/0.7)
θ = 60.8⁰
Thus, the speed of the second ball after impact is 1.43 m/s at 60.8⁰.
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A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.
The tangential speed of the mass is 2.85 m/s. Calculate the angle between the string and the vertical.
Answers
The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have
• net horizontal force
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]
where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.
As shown in the diagram, we can see that
[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]
where [tex]r=1.69\,\rm m[/tex], so that
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]
The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have
• net vertical force
[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]
Solve for [tex]\theta[/tex] :
[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]
Complete the square:
[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]
Plugging in the known quantities, we end up with
[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]
The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with
[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]
Due to continental drift, Africa and South America are moving away from each other at a rate of 4 centimeters per year. The two coasts are currently separated by 5,000 km. Assuming this drift rate is constant, how long ago were the coasts touching. Answer in millions of years
Answers
The coasts of Africa and South America were touched 125 million years ago due to continental drift.
The time can be computed from the ratio of distance and speed.
Given:
Distance = 5, 000,000
Speed = 4 cm/year = 0.04 m/year
The time is computed as:
Time = Distance / Rate
Time = 5,000,000 meters / 0.04 meters per year
Time = 125,000,000 years
Hence, the coasts of Africa and South America were touched 125 million years ago due to continental drift.
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Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel.
During the process, 30 kJ of heat is transferred to the water, and 5 kJ of heat is lost to the
surrounding air. The paddle-wheel work amounts to 500 N · m. Determine the final energy of
the system if its initial energy is 10 kJ.
Answers
Heat transfer in a closed system is the addition of changes in internal energy and the total amount of work done by it. The final energy of the system is 35.5kJ.
What is heat transfer?Heat transfer is the transfer of heat energy due to temperature differences.
The paddle-wheel paintings are quantities of workdone, 500 N.m or 0.5kJ.
The preliminary (initial) power of the device is 10 kJ.
Total warmness transferred in the course of the method is 30 kJ
Total warmness misplaced in the course of the method to the encompassing air is 5 kJ.
The energy of the system is given as:
The energy of the system = Energy in - Energy out
The energy of the system = Initial energy + Energy transferred + Work done - Energy lost
Energy of the system = 10 + 30 + 0.5 - 5 kJ
Energy of the system = 35.5 kJ
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A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.
(a)What is the maximum potential difference across the resistor (in V)?
(b)What is the maximum current through the resistor (in A)?
(c)What is the rms current through the resistor (in A)?
(d)What is the average power dissipated by the resistor (in W)?
Answers
(a) The peak voltage across the resistor is 339.41 V.
(b) The maximum current through the resistor is 0.23 A.
(c) The rms current through the resistor is 0.16 A.
(d) The average power dissipated by the resistor is 38.4 W.
Peak voltage or maximum potential differenceVrms = 0.7071V₀
where;
V₀ is peak voltageV₀ = Vrms/0.7071
V₀ = 240/0.7071
V₀ = 339.41 V
rms current through the resistorI(rms) = V(rms)/R
I(rms) = (240)/(1,540)
I(rms) = 0.16 A
maximum current through the resistorI₀ = I(rms)/0.7071
I₀ = (0.16)/0.7071
I₀ = 0.23 A
Average power dissipated by the resistorP = I(rms) x V(rms)
P = 0.16 x 240
P = 38.4 W
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In an experiment, the __________ is what researchers measure and expect to change as a result of manipulation.
Answers
Answer:
dependent variable
Explanation:
How much did mary pay in social securty taxes
Answers
The amount that Mary will pay in social security taxes will be $1550.
How to compute the value?It should be noted that the current tax for the social security is 6.2% for the employer and 6.2% for the employee.
In this case, let's assume that the income for Mary is $25000, the amount that Mary will pay will be:
= 6.2% × $25000
= 0.062 × $25000
= $1550
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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced. What is the upward force exerted on the board by the support?
Answers
The upward force exerted on the board by the support is mathematically given as
Fu= 764.8 N
What is the upward force exerted on the board by the support?Generally, the equation for is mathematically given as
Considering that the Net Force on the system is null
The weight of the children plus the weight of the board equals the upward force imposed on the support.
The upward force
Fu= 440 + 272 + 52.8 N
Fu= 764.8 N
In conclusion, he upward force exerted on the board by the support
Fu= 764.8 N
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Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.56 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.2 mm. If the floor is carpeted, this stopping distance is increased to about 1.3 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. Answer the following:
1.hardwood floor magnitude
2.hardwood floor duration
3.carpeted floor magnitude
4.carpeted floor duration
Answers
This question is health related and seeks to compute the details surrounding a traumatic head injury. See the explanation below.
What is a traumatic head injury?Traumatic brain injury (TBI), a form of acquired brain injury, occurs when a sudden trauma causes damage to the brain.
TBI can result when the head suddenly and violently hits an object, or when an object pierces the skull and enters brain tissue
What are the details surrounding the head injury?Final velocity before hitting the ground
V² = 2g*0.56
V² = 2 * 9.8 * 0.56
V = √10.976
V = 3.312 m/s
*** Note that g which is the acceleration of gravity is about 9.8 meters.
1a) Where there is a hardwood floor
-u^2 = -2as
a = 3.312^2/2*0.0019
a = 1,042.08 m/s^2
1b) Duration
t = u/a
= 3.312/1,042.08
t = 3.18 ms
2a) Where there is a carpet on the floor
a = 3.312^2/2*0.013
a = 713.01 m/s^2
t = u/a
= 3.312/713.01
t = 4.65 ms
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In a game at a carnival, a contestant rolls a ball up the slope with an initial speed vi. The object of the game is to roll the ball in such a way that it will get “stuck” in the depression at B and not return back down the slope. This will happen if the ball’s speed when it gets to point A is essentially zero. (The speed of the ball at point A really has to be greater than zero in order for the ball to make it past point A, but
the speed at point A must be greater than zero only by an arbitrarily small amount, so that we can say that the condition for the ball not to return is essentially that the speed at A must be zero.) Assuming that the ball rolls without slipping and that energy losses due to friction are negligible, find the initial speed vi required to make the
speed of the ball at point A zero. Let the mass of the ball be called M and let the radius of the ball be called R. Treat the ball as a solid sphere.
Answers
Initial speed of ball will be 2.7m/s.
When a body is rotating about an axis, then it has kinetic energy.
And this energy is called rotational kinetic energy.
It is given as - R.K.E. = 1/2 Iω²
And if a ball is rolling without slipping.
Then the moment of inertia of the solid ball is written as -
I = 25MR²
Vi = Rω
Here it is given in the problem that-
height(h) = 0.53m
Now by the conservation of energy we can write the equation as -
1/2MVi² + 1/2Iω² = Mgh
so that -
(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²
= gh(7/10)Vi² = 9.8 × 0.53
Vi = 2.7 m/s
So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.
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How is Behaviorism and Psychoanalysis theories are different?
Answers
Answer:
Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.
Explanation:
Answer: Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.
Explanation:
a metal ion (x) with a charge of 3= is attracted to a normal ion (z) with a charge of 4- which of these formulas represents the resulting comppound
Answers
If the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.
What is the formula of a compound?The formula of an ionic compound is determined by the charges of the ions that make up the compound. Recall that when the compound is formed, the ions exchange valences.
Hence, is the metal ion has a 3+ charge and the nonmetal ion has a 4- charge, the formula of the compound is X4Z3.
Missing parts:
A metal ion (X) with a charge of 3+ is attracted to a nonmetal ion (Z) with a charge of 4-. Which of these formulas represents the resulting compound.
A. 7XZ
B. X3Z4
C. X4Z3
D. 4X3Z
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An object has a weight of 9 n when it is in air and 7.2 n when it is submerged into water. what is the specific gravity of the object’s material?
Answers
The specific gravity of the object’s material is 5.09.
To calculate the specific gravity of the object:Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water
Buoyant Force in water(Fb) = density of water x g x volume of the body(Vb)
1.8 = 1000 x 9.81 x Vb
Vb = 1.8/9810 cubic meter
Now, in the air;
Weight of body = mg = 9 N
Mass of body,m = 9/9.81 Kg
So,
Density of body = m/ Vb
= 9/9.81 ÷ 1.8/9810
= 5094.44 kg per cubic meter
The specific gravity of body = density of body ÷ density of water
= 5094.44 ÷ 1000
= 5.09
Therefore, Specific gravity of body = 5.09
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A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs of the same constant K= 250 N/m, then the block is pulled a distance A and released. What's the speed V and angular velocity w of the block as it passes through the equilibrium when the two springs are arranged in:
a) parallel
b) series
Answers
a.
i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/sii. the angular velocity when the two springs are in parallel is 7.07 rad/sb.
i. the speed of the block of mass when the springs are connected in series is 11.2 A m/sii. the angular velocity when the two springs are in series is 11.2 rad/sa. i. How to calculate the velocity of the mass when the springs are connected in parallel?Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k
= 2k
= 2 × 250 N/m
= 500 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where
k' = equivalent spring constant in parallel = 500 N/m, A = maximum displacement of spring, x = equilibrium position = 0 m, M = mass of block = 10 kg and v = speed of block at equilibrium positionMaking v subject of the formula, we have
v = √[k'(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k'(A² - x²)/M]
v = √[500 N/m(A² - (0)²)/10]
v = √[50 N/m(A² - 0)]
v = [√50]A m/s
v = [5√2] A m/s
v = 7.07 A m/s
So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
ii. The angular velocity of mass when the springs are in parallel
Since velocity of spring v = ω√(A² - x²) where
ω = angular velocity of spring, A = maximum displacement of spring and x = equilbrium position of spring = 0 mMaking ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 7.07 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 7.07 A m/s/√(A² - 0²)
ω = 7.07 A m/s/√(A² - 0)
ω = 7.07 A m/s/√A²
ω = 7.07 A m/s/A m
ω = 7.07 rad/s
So, the angular velocity when the two springs are in parallel is 7.07 rad/s
b. i. How to calculate the velocity of the mass when the springs are connected in series?Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k
= 2/k
⇒ k" = k/2
k" = 250 N/m ÷ 2
= 125 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where
k" = equivalent spring constant in series = 125 N/m, A = maximum displacement of spring, x = equilibrium position = 0 m, M = mass of block = 10 kg and v' = speed of block at equilibrium positionMaking v subject of the formula, we have
v = √[k"(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k"(A² - x²)/M]
v = √[125 N/m(A² - (0)²)/10]
v = √[125 N/m(A² - 0)]
v = [√125]A m/s
v = [5√5] A m/s
v = 11.2 A m/s
So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s
ii. The angular velocity of the mass when the springs are in series
Since velocity of spring v = ω√(A² - x²) where
ω = angular velocity of spring, A = maximum displacement of spring and x = equilbrium position of spring = 0 mMaking ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 11.2 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 11.2 A m/s/√(A² - 0²)
ω = 11.2 A m/s/√(A² - 0)
ω = 11.2 A m/s/√A²
ω = 11.2 A m/s/A m
ω = 11.2 rad/s
So, the angular velocity when the two springs are in series is 11.2 rad/s
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A charge of 5.4 C experiences a force of 25.0 in an electric field. What is the strength of electric field at that point ? If the 5.4 C is replaced by a charge of half the magnitude, 2.7 C, how much force would this charge experience at that point ? Remember to identity all of your data, write the equation, and show your work. (5 points)
Answers
The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and 12.38 N.
What is the electric field strength?The electric field strength is defined as the ratio of electric force to charge.
Given data;
q₁ = 5.4 C
F₁ is the electric force in case1
E is the electric field =?
F₂ is the electric force in case 2
q₂ is the charge 2
The strength of the electric field at that point is;
F₁=Eq₁
E₁=F/q₁
E₁=25.0 N / 5.4 C
E₁=4.587 N/C
The force would this charge experience at that point when the charge is 2.7 C;
F₂=Eq₂
F₂=4.587 N/C × 2.7 C
F₂ = 12.38 N
Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and 12.38 N.
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Assume the three resistances (R1, R2, R3) in the problem 4 are now connected in parallel. (A) What is the total resistance of the parallel resistors? (B) What is the current in the overall circuit? (C) What is the current through each resistance?
Answers
(A)The total resistance of the parallel resistors is: [tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]
(B) The current in the overall circuit is: [tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]
(C) The current through each resistance is as follows:
[tex]I_{1} = \frac{V}{R_{1}}[/tex][tex]I_{2} = \frac{V}{R_{2}}[/tex][tex]I_{3} = \frac{V}{R_{3}}[/tex]What is the the total resistance for resistors in parallel?For the three resistances connected in parallel, R1, R2, R3 , the total resistance, [tex]R_{T}[/tex] is calculated as follows:
[tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]
The current in the overall circuit is calculated using the formula:
[tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]
The current through each resistance is given as follows;
Current through R1, [tex]I_{1} = \frac{V}{R_{1}}[/tex]
Current through R2; [tex]I_{2} = \frac{V}{R_{2}}[/tex]
Current through R3; [tex]I_{3} = \frac{V}{R_{3}}[/tex]
In conclusion, the voltage across resistances in parallel is the same but the current varies.
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What is the force between two balloons with a negative charge of 1.6 × 10−10 C if the balloons are 0.05 m apart?
Answers
Considering the Coulomb's Law, the force between the two ballons is 9.216×10⁻⁸ N.
Coulomb's LawCharged bodies experience a force of attraction or repulsion on approach.
From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.
From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
[tex]F=k\frac{Qq}{d^{2} }[/tex]
where:
F is the electrical force of attraction or repulsion. It is measured in Newtons (N).Q and q are the values of the two point charges. They are measured in Coulombs (C).d is the value of the distance that separates them. It is measured in meters (m).K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ [tex]\frac{Nm^{2} }{C^{2} }[/tex].The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
Force between two balloonsIn this case, you know that two balloons have a negative charge of 1.6×10⁻¹⁰ C and the balloons are 0.05 m apart.
Replacing in the Coulomb's Law, you get:
[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } \frac{(-1.6x10^{-10} C)x(-1.6x10^{-10} C)}{(0.05 m)^{2} }[/tex]
Solving:
[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } \frac{2.56x10^{-20} C^{2} }{(0.05 m)^{2} }[/tex]
[tex]F=9x10^{9} \frac{Nm^{2} }{C^{2} } 1.024x10^{-17}\frac{ C^{2} }{m^{2} }[/tex]
F= 9.216×10⁻⁸ N
Finally, the force between the two ballons is 9.216×10⁻⁸ N.
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What is the equivalent resistance of a circuit that contains three 10.0 Ω
resistors connected in series to a 6.0 V battery?
Ο A. 0.600 Ω
Ο Β. 60.0 Ω
Ο C. 16.0 Ω
Ο D. 30.0 Ω
Answers
Question :
What is the equivalent resistance of a circuit that contains three 10.0 Ω resistors connected in series to a 6.0 V battery?Options given :
A. 0.600 ΩΒ. 60.0 ΩC. 16.0 ΩD. 30.0 ΩSolution :
As we know that,
When a circuit is connected in series resistance is calculated by :
Rs = R1 + R2 + R3 .... RnHere,
R is resistanceAs a circuit contains three resistors. So R1 , R2 and R3 would be :
R1 = 10 ΩR2 = 10 ΩR3 = 10 ΩNow,
>> Rs = 10 + 10 + 10
>> Rs = 30 Ω
Therefore,
Option D is correct.The equivalent resistance in the circuit is 30.0 Ω. The correct option is D.
What is Ohm's law?Ohm's law is a fundamental principle of electric circuits that describes the relationship between electric current, voltage, and resistance. It states that the electric current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.
Mathematically,
V = IR
Where V = is the voltage across the conductor in volts (V),
I = is the electric current flowing through the conductor in amperes (A),
R =is the resistance of the conductor in ohms (Ω).
This relationship can also be expressed in terms of power, using the formula:
P = VI
Where P = is the power in watts (W),
V = is the voltage in volts (V),
I = is the current in amperes (A).
Ohm's law is named after the German physicist Georg Simon Ohm, who first published it in 1827 as a result of his experiments with electric circuits. Ohm's law is widely used in electrical engineering and is essential for understanding and designing electronic devices and circuits.
Here in this question,
The equivalent resistance of resistors in series is the sum of their individual resistances.
Now, the equivalent resistance of three 10.0 Ω resistors connected in series is:
Req = R1 + R2 + R3
Req= 10.0 Ω + 10.0 Ω + 10.0 Ω
Req= 30.0 Ω
Using Ohm's law, we can find the current flowing through the circuit:
I = V/R
= 6.0 V/30.0 Ω
I = 0.2 A
Therefore, the equivalent resistance of the circuit is 30.0 Ω (Option D).
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If the coefficient of kinetic friction between a 24-kg crate and the floor is 0.25 what horizontal force is required to move the crate at a steady speed across the floor?
Answers
The horizontal force required to move the crate at a steady speed across the floor will be 58.86 N.
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
The given data in the problem is;
Mass,m = 24 kg
The coefficient of kinetic friction, μ=0.25
The friction force is defined as the product of the coefficient of friction and normal reaction.
f=μN
If there is no other force, the horizontal force is equal to the friction force;
F=f
The horizontal force required to move the crate at a steady speed across the floor is found as;
F = μN
The normal reaction force is;
N=W
N=mg
F = μmg
F=0.25 × 24-kg × 9.81 m/s²
F = 58.86 N
Hence horizontal force is required to move the crate at a steady speed across the floor will be 58.86 N.
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A Skier starting from rest rolls down a 35.0 m ramp. When he arrives at the bottom of the ramp his speed is 10.43 m/s. (a) Determine the magnitude of his acceleration, assumed to be constant, (b) If the ramp is inclined at 60.0 degrees with respect to the ground, what is the component of his acceleration that is parallel to the ground. (c) Provide me with a free body diagram, one for the 35 m ramp situation. Also, provide a nice sketch of the skier.
Answers
We obtain the magnitude of the acceleration as 1.55 m/s^2 while its component parallel to the ground is 0.775 m/s^2.
What is acceleration?The acceleration refers to change of speed with time. From the question;
distance covered = 35.0 m final velocity = 10.43 m/sIt the follows that;
v^2 = u^2 + 2as
And;
u = 0m/s owing to the fact that motion started from rest
v^2 = 2as
a = v^2/2s
a = (10.43 m/s)^2/2 * 35.0 m
a = 1.55 m/s^2
For the acceleration component which is parallel to the ground;
ax = a sin θ
ax = 1.55 m/s^2 cos 60
ax = 0.775 m/s^2
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How do hydrogen bonds affect boiling points?
Answers
Boiling points are raised by hydrogen bonds because they make different molecules desire to "attach" to one another, which requires more energy to do so. In water, for instance, the hydrogen proton is in a state that resembles ionization because the connections between oxygen and hydrogen, while covalent, are strongly polar. The oxygen also receives a partial negative charge. Therefore, hydrogen bonds are formed when the electro-positive H in one molecule is strongly electrostatically attracted to the electro-negative O in nearby molecules. Despite being weak links, they are powerful enough to significantly alter the liquid's characteristics.
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The speed of a bus is reduced uniformly from 20 m/s to 10 m/s while traveling 60 m. (a)
Compute the acceleration. (b) How much farther will the bus travel before coming to rest,
provided the acceleration remains constant? (c) Draw a diagram showing the motion from start to
finish of the bus.
Answers
The bus will travel a further 20 m before coming to rest.
What is acceleration?The term acceleration has to do with a change in velocity with time. Now we have;
u = 20 m/s
v = 10 m/s
s = 60 m
Now;
v^2 = u^2 -2as
v^2 - u^2 = -2as
(10)^2 - (20)^2 = - 2 * a * 60
a = (10)^2 - (20)^2/ - 2 * 60
a = 100 - 400/ - 2 * 60
a = 2.5 m/s^2
At that time;
v = 0 m/s
u = 20 m/s
a = 2.5 m/s^2
s = ?
Hence;
u^2 = 2as
s = u^2/2a
s = (20)^2/ 2 * 2.5
s = 400/5
s = 80 m
Hence, the bus will travel a further 20 m before coming to rest.
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