Assuming players stick together immediately after the collision, their speed will be 1.3206 m/sec, and - ve indicates their direction.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of 1st player= 101.0 kg
(u₁) is the initial velocity of 1st player (u₁) = 4.10 m/s
(m₂) is the mass of 2nd player = 117 kg
(u₂) is the initial velocity of 2nd player= -6 m/s
(v) is the velocity after collision =.?
According to the law of conservation of momentum;
Momentum before collision =Momentum after collision
m₁u₁+m₂u₂=(m₁+m₂)V
101.0 kg × 4.10 m/sec + 117 kg × (-6 m/sec) = (101.0 kg + 117 kg) × V
V= - 1.3206 m/sec
Hence, the velocity of players just after impact, if they cling together, will be - 1.3206 m/sec, and - ve shows the direction.
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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.700 m from the pivot.
Calculate the tension in the cable.
The tension in the cable is 23.2 N
What is the tension in the string?The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.
Tcosθ, is acting perpendicularly, Tcosθ = 0
Taking moments about the pivot:
Tsinθ * 2.2 = 4 * 9.8 * 0.7
Solving for θ;
θ = tan⁻¹(1.4/2.2) = 32.5°
T = 27.44/(sin 32.5 * 2.2)
T = 23.2 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
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Sally and Sam are in a spaceship that comes to within 18,000 km of the asteroid Ceres. Determine the force Sally experiences, in N, due to the presence of the asteroid. The mass of the asteroid is 8.7 1020 kg and the mass of Sally is 80 kg. For calculation purposes, assume the two objects to be point masses.
The force Sally experiences, in N, due to the presence of the asteroid is 0.0143 N.
Force experienced by Sally
F = Gm1m2/R²
where;
G is universal gravitation constantm1 is mass of asteroidm2 is mass of SallyR is the distance between Sally and SamF = (6.67 x 10⁻¹¹ x 80 x 8.7 x 10²⁰)/(18,000,000)²
F = 0.0143 N
Thus, the force Sally experiences, in N, due to the presence of the asteroid is 0.0143 N.
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Which characteristic of a sound wave is directly related to the loudness of the sound?
The characteristic of a sound wave that is directly related to the loudness of the sound is amplitude.
What is sound waves?Sound wave is a type of wave that is produced by the vibration of objects which can be heard only if it's of high frequency.
The properties of sound waves include:
loudness or volumeamplitude,quality.The characteristic of a sound wave that is directly related to the loudness of the sound is amplitude. That is, a larger amplitude means a louder sound, and a smaller amplitude means a softer sound.
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Set local ground level to 700 ft, and record the inches of mercury.
What is the altimeter reading in inches of mercury (from the Kollsman window)?
The main way in which an altimeter measures the altitude of an object is by calculating the location's air pressure.
What is an Altimeter?This refers to the instrument that is used to measure the altitude of an object when it is at a fixed level.
Hence, we can see that an altimeter should NOT be confused with a barometer, because although they both measure pressure, a barometer calculates the change in air pressure and elevation based on available weather.
Please note that your question is incomplete so I gave you a general overview to help you get a better understanding of the concept.
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A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 112 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 37.8°.
How wide is the river?
y = _____________ m
y = 11.3380m
What is Young's double-slit experiment?The double-slit experiment is an experiment, that shows that light has both a wave nature or characteristic and a particle nature or characteristic, and that these natures are inseparable.
So, light is said to have wave–particle duality rather than be only a wave or only a particle. The same is true for electrons and other quantum particles.
According to the question,
The relative to angle θ, its adjacent side has length x and its opposite side is equal to width of the river, y;
tanθ = [tex]\frac{y}{d}[/tex] = y = dtanθ
y =( 112m) tan (37.8° )
y ≈ 11.3380m
The width of river is 11.3380m
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Adults sometimes lump all "screen time" together—social media, music videos, games and so on. Is this valid, or are there distinct differences among different types?
Answer:
it’s all time on the Screen so it should be counted as one thing they are different things that you have different timelines for each category
Since everything is happening at once on the screen, even though there are distinct events with separate timings for each category, it should all be considered one item.
What are social media ?Websites and programs that place a high value on community involvement, content sharing, collaboration, and social interaction are referred to as social media. Social media is used by people to interact and communicate with their friends, family, and other groups.
Why is social media so important?It is a helpful tool for talking with individuals locally and internationally as well as for generating, sharing, and spreading information. Social media has the power to influence consumer buying decisions via reviews, marketing, and marketing tactics.
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How is Behaviorism and Psychoanalysis theories are similar?
Answer:
One of the key similarities between both the psychoanalytical theory and the behavioral theory is that both are by and large, deterministic.
A rocket uses 400.0 J of chemical potential energy stored in the fuel while shooting the 0.55 kg rocket straight up
into the air. The rocket reaches a height of 23 m. What was the efficiency of the rocket in transforming the chemical
potential energy of the fuel into gravitational potential energy?
Select one:
O a. 25%
Ob. 35%
O c. 31%
O d. 29%
Answer:
Approximately [tex]31\%[/tex] assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
Explanation:
Consider an object of mass [tex]m[/tex] in a uniform gravitational field of strength [tex]g[/tex]. If the height of that object increased by [tex]\Delta h[/tex], the gravitational potential energy of that object would increase by [tex]m\, g\, \Delta h[/tex].
In this question, the mass of the rocket is given to be [tex]m = 0.55\; {\rm kg}[/tex]. Assume that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex]. The rocket has gained a height of [tex]\Delta h = 23\; {\rm m}[/tex]. Thus, the gravitational potential energy of this rocket would have increased by:
[tex]\begin{aligned} m\, g\, \Delta h &= 0.55\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 23\; {\rm m} \\ &\approx 124.1\; {\rm J} \end{aligned}[/tex].
In other words, the useful energy output from the combustion of the rocket fuel was approximately [tex]124.1\; {\rm J}[/tex].
The energy input to this rocket was given to be [tex]400.0\; {\rm J}[/tex]. Thus, the efficiency of the energy conversion would be:
[tex]\begin{aligned} \text{efficiency} &= \frac{(\text{useful energy out}\text{put})}{(\text{energy in}\text{put})} \times 100\% \\ &\approx \frac{124.1\; {\rm J}}{400.0\; {\rm J}} \times 100\% \\ &\approx 31\%\end{aligned}[/tex].
What is resistance force?
A. The force opposing the effort
B. The force exerted by the machine
C. The work put into a machine
D. A force that improved on your effort?
Answer asap
Answer:
A
Explanation:
resisting is doing something that goes against what something or someone is doing therefore the resisting force is the opposing force
Annie is considering two different coach journeys from Bristol to Birmingham.
Journey A takes 2 hours and does not stop. The shortest distance by road between the wo coach stations is 143 km, most of which s on the motorway where the speed limit is 31m/s (70 mph).
Journey B costs only half as much, but it takes 3 hours 15 minutes and stops at 5 other towns on the way. This journey is 160 km long, and most of it is on country roads with a speed limit of 22 m/s (50 mph). The speed limit in towns the coach passes through is 13 m/s (30 mph).
Calculate the average speed for each journey in m/s. Think carefully about the units you need to use in your calculations.
The average speed in the first case is 19.86 m/s and the average speed in the second case is 13.67 m/s.
The formula of average speed is
average speed = total distance / total time
In first case
total distance is 143 km which is also 143000 m
and total time is 2 hours which is also 7200 seconds
so average speed = ( 143000 / 7200 ) m/s
average speed = 19.86 m/s
In second case
total distance is 160 km which is also 160000 m
and total time is 3 hours and 15 minutes which is also 11700 seconds
so average speed = ( 160000 / 11700 ) m/s
average speed = 13.67 m/s
So to conclude with we have find the the average speed in both cases by dividing total distance by total time and in first case we have got our answer as 19.86 m/s and in second case we have got 13.67 m/s.
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Lesson 2 History of Physical Science
Write an expository essay explaining how science builds on itself. Use at least two specific examples from this lesson.
Science builds on itself because the process of science is iterative and not predetermined.
How do science builds on itself?Science is said to build on itself because the process of science is iterative. This means that science does not have a linear path but circles back on itself so that useful ideas are built upon and used to learn even more about the natural world.
For instance, the discovery of inheritance that could be passed from parent to their offspring by Gregor Mendel.
Many more scientist has worked subsequently on chromosome, gene and DNA.
They have continued to deepen and extend our understanding of genes, how they are controlled, how patterns of control themselves are inherited, and how they produce the physical traits that pass from generation to generation.
Science is said to build on itself because the process of science is not predetermined. This means that carrying out an experiment on an idea would lead to the discovery of a new idea altogether.
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(a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.
N
(b) Calculate the tension in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.13. The strand sags at an angle of 10.0° below the horizontal.
N
Compare this with the tension in the vertical strand (find their ratio).
(tension in horizontal strand / tension in vertical strand)
a)The tension in the vertical strand will be 0.00049 N.
b))The tension in the horizontal strand will be 0.001284 N.
c)The ratio of tension in the horizontal strand to the tension in the vertical strand will be 2.62.
What is tension force?
The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.
Given data;
Mass of spiderweb = 5.00 ✕ 10⁻⁵ kg
The tension in a horizontal strand of a spiderweb, Tₓ
The tension in a vertical strand of a spiderweb, Tₐ
The angle at which the strand sags, Θ = 10.0°
R is the ratio of tension in the horizontal strand to the tension in the vertical strand
a)
From the FBD,
The tension in the vertical strand is equal to the weight of the spider;
Tₐ = W
Tₐ = mg
Tₐ= 5 x 10⁻⁵ kg x 9.8 m / s²
Tₐ= 0.00049 N.
(b)
The super's total vertical forces are zero since it is not moving the value of the tension in the horizontal strand;
Tₓ = mg/( 2sinΘ)
Tₓ = 5 x 10⁻⁵ kg x 9.8 m /s² / ( 2 sin 11° )
Tₓ= 0.001284 N.
c)
R is the tension in the vertical strand divided by the tension in the horizontal strand.
R= Tₐ / Tₓ
R= 0.001284 N / 0.00049 N
R= 2.62
Hence, the tension in a vertical strand of a spiderweb, the tension in a horizontal strand of a spiderweb, and the ratio of tension in the horizontal strand to the tension in the vertical strand will be 0.00049 N, 0.001284 N, and 2.62 respectively.
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What is the power of a motor that transfers 550 J of energy in 11s?
Answer:
The answer is 50 watts
Explanation:
The formula of power is work done over elapsed time
P= W/t
P= 550 J / 11 sec
P= 50 watts
A house is advertised as having 1580 square feet under roof. What is the area of this house in square meters?
The area of the house will be 146.78 in square meters
What we have given in the question is that there is one house whose area is measured in square feet which is 1580 square feet, and we have to convert that area in square meter.
Now, To convert the area in square meter we must divide the area given in square feet by a constant which is 10.764.
Area in square meter = Area in square feet / 10.764
Area in square meter = 1580 / 10.764 square meter
Area in square meter = 146.78 square mete
Thus we conclude that You must divide the area given in square feet by 10.764 to convert Square feet to Square meters and vice versa for the opposite and after calculation we found that area which is 1580 square feet will be 146.78 in magnitude in square meters
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Many years ago, a scientist discovered an animal P and studied its characteristics. Some of the characteristics of animal P are as follows:
A Lays egg
B has four legs
C Can swim in the water
D Can produce milk to feed the young.
Which of the above characteristics made it difficult to classify animal P as a mammal?
Answer:
A
Explanation:
most mammals are viviparous instead of oviparous. oviparous means laying eggs
How is Behaviorism and Psychoanalysis theories are similar and different?
A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads. Find the amount of electric charge that resides on each capacitor once it is fully charged.
The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
Total capacitance of the circuit
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
Charge in each capacitorQ = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
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Find the magnitude of this
vector:
174 m
N
188.4 m
HELP FAST
Answer:
195.168 m
Explanation:
To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse
Pythagorean Theorem:
[tex]a^2+b^2=c^2[/tex]
Plug values in
[tex]88.4^2+174^2=c^2[/tex]
Simplify
[tex]7814.56+30276=c^2[/tex]
Add the two values
[tex]38090.56=c^2[/tex]
Take the square root of both sides
[tex]195.168\approx195.168[/tex]
A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 23 m/s. The third piece has three times the mass of the other two pieces.
Write an expression for the speed of the larger piece, that is in terms of only the variable V.
What is the numeric value for the speed of the larger piece, in meters per second?
At what angle does the largest piece travel with respect to the -y axis, in degrees?
(a) The expression for the speed of the larger piece is [tex]\frac{v\sqrt{2} }{3}[/tex].
(b) The numeric value for the speed of the larger piece is 10.84 m/s.
(c) The direction of the larger piece is 45⁰ with respect to y - axis.
Expression for the speed of the larger piece
Apply the principle of conservation of linear momentum;
m₁v₁ + m₂v₂ + m₃v₃ = 0
where;
m₁ is mass of first piecem₂ is mass of second piecem₃ is mass of third piece = three times the mass of other piece = 3mmv.i + mv.j + (3m)v₃ = 0
(3m)v₃ = -mv.i - mv.j
3v₃ = - v.i - v.j
[tex]v_3 = \frac{-v_i \ - \ v_j}{3} = \frac{v\ (\sqrt{(-1)^2 + (-1)^2} )}{3} = \frac{v\ \sqrt{2} }{3}[/tex]
Numeric value for the speed of the larger piece[tex]v_3 = \frac{23 \times \sqrt{2} }{3} \\\\v_3 = 10.84 \ m/s[/tex]
Direction of the larger piecetan θ = vj/vi
tan θ = 1/1
tan θ = 1
θ = arc tan(1)
θ = 45⁰
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How do i solve for the point of equigravity between two planets, given their mass ratio and the distance between them ?
The equigravity between two planets is determined from the product and their masses and square of distance between them.
Gravitational force between two planetsThe gravitational force between two planets is calculated as follows;
F = GM₁M₂/R²
where;
G is universal gravitation constantM₁ is mass of first planetM₂ is mass of the second planetR is the distance between the two planetsUse the mass ratio of the two planets to determine their individual masses.
Thus, the equigravity between two planets is determined from the product and their masses and square of distance between them.
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A car drives 16 miles south and then 12 miles west. What is the magnitude of the car's displacement?
O4 miles
16 miles
20 miles
28 miles
Answer:
20 miles
Explanation:
Use pythagorean theorem
d^2 = 16^2 + 12^2
to find displacement, d = 20 miles
A supersonic airplane is flying horizontally at a speed of 2610 km/h.
1. What is the centripetal acceleration of the airplane, if it turns from North to East on a circular path with a radius of 80.5 km?
2. How much time does the turn take?
3. How much distance does the airplane cover during the turn?
Answer:
Centripetal acceleration of this aircraft: approximately [tex]6.52\; {\rm m \cdot s^{-2}}[/tex].
Distance covered during the turn: approximately [tex]63.2\; {\rm km}[/tex].
Time required for the turn: approximately [tex]0.0242\; \text{hours}[/tex] (approximately [tex]87.2\; {\rm s}[/tex].)
Explanation:
Convert velocity and radius to standard units:
[tex]\begin{aligned}v &= 2610\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
[tex]\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}[/tex].
Hence, the centripetal acceleration of this aircraft:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(725\; {\rm m\cdot s^{-1}})^{2} }{8.05 \times 10^{4}\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
The trajectory of the turn is an arc with a radius of [tex]r = 80.5\; {\rm km}[/tex] and a central angle of [tex]\theta = 90^{\circ} = (\pi / 4)[/tex]. The length of this arc would be:
[tex]\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}[/tex].
The time required to travel [tex]63.2\; {\rm km}[/tex] at a speed of [tex]2610\; {\rm km \cdot h^{-1}}[/tex] would be:
[tex]\begin{aligned}t &= \frac{s}{v} \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^{-1}}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} \times \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}[/tex].
An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).
An object labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the object is displaced by a distance A to the right of its equilibrium position.
The object is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a)
At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
N
(b)
At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
Toward the equilibrium position (i.e., to the left in the figure).
Away from the equilibrium position (i.e., to the right in the figure).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.
Answer:
See below
Explanation:
a) Spring force at release = k * d = 132 N/m * .120 m = 15.84 N
b) F = ma
15.84 = (.660 kg)(a) a = 24 m/s^2
c) Toward the left ....the object is accelerated to the left
A. The magnitude of the spring force acting upon the object at the instant of release is approximately 15.84 N.
B. With the moment of release, the object is moving with an acceleration of about [tex]\rm -24 m/s^2[/tex].
C. The correct option is toward the equilibrium position (i.e., to the left in the figure).
Let us calculate the magnitude of the spring force, the acceleration of the object and the direction of the vector at the time of release.
Given:
Mass of the object (m) = 0.660 kg
Force constant of the spring (k) = 132 N/m
Displacement from equilibrium (x) = 0.120 m
Magnitude of the spring force [tex]\rm (F_s_p_r_i_n_g):[/tex]
To determine the spring force at the time of release, we can apply Hooke's law. According to Hooke's law, the force of a spring depends on how far it has moved from its equilibrium position. Hooke's law is expressed as:
[tex]\rm F_s_p_r_i_n_g[/tex] = -k * x
where
[tex]\rm F_s_p_r_i_n_g[/tex] is the spring force,
k is the force constant of the spring, and
x is the displacement from the equilibrium position.
When we put the values, we get:
[tex]\rm F_s_p_r_i_n_g[/tex] = -132 N/m * 0.120 m
[tex]\rm F_s_p_r_i_n_g[/tex] = -15.84 N
Therefore, the magnitude of the spring force acting upon the object at the instant of release is approximately 15.84 N.
The net force acting on the object at the time of release determines its acceleration. The spring force and any additional external forces acting on the object add up to form the net force. In this case the spring force is the only force acting on the object.
According to Newton's second law of motion, the acceleration of an object is inversely related to its mass and directly proportional to the total force exerted on it. Acceleration is calculated as:
a = [tex]\rm F_n_e_t[/tex] / m
where
a is the acceleration,
[tex]\rm F_n_e_t[/tex] is the net force, and
m is the mass of the object.
Noting that the mass (M) is 0.660 kg and the net force (-15.84 N) is equal to the force of the spring, we can plug the following numbers into the formula:
a = (-15.84 N) / 0.660 kg
a = [tex]\rm -24 m/s^2[/tex]
Therefore, with the moment of release, the object is moving with an acceleration of about [tex]\rm -24 m/s^2[/tex].
The direction of the acceleration vector is left in the diagram, or towards the equilibrium position. This is because the negative sign of the acceleration value indicates that the acceleration is moving in the direction opposite to that in which the equilibrium position has moved. Since the signal indicates that the object first moved to the right, acceleration will be applied to the left in an attempt to bring the object back to its equilibrium position.
So, the correct option is toward the equilibrium position (i.e., to the left in the figure).
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An open bed truck is moving at 20 m/s and comes to a stop. A 100 kg crate sitting in the back of the truck remains at rest in the truck bed. How much work was done by friction to keep the crate from moving
as the truck came to a stop?
[?] J
The work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.
Work done by friction to prevent the crate from moving
The work done is determined by applying the principle of conservation of energy.
W = K.E
W = ¹/₂mv²
where;
m is mass of cratev is speed of the crate with respect to the truck = 20 m/sW = ¹/₂(100)(20²)
W = 20,000 J
Thus, the work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.
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A merry-go-round has a small box placed on it at a distance R' from its axis of
rotation. Initially (at time = 0), the merry-go-round is stationary. At time t = 0.0, a
constant torque of 6 Nm is applied to it, causing it to start rotating. As the merry-
go-round spins faster, a point comes when the small box slides off of it. Given the
following data, calculate the time after it starts spinning, that the box will slip off
the disk. Assume that the merry-go-round is a solid disc. Write your answer in
seconds.
Mass of the merry-go-round (disk) = 30 kg.
Radius of the merry-go-round = 3.1 m
Distance from the axis where the box is placed = R = 1.4 m
Coefficient of friction between merry-go-round and box: Static = 0.5, kinetic
The time after it starts spinning, that the box will slip off is 0.1 s.
Apply the principle of conservation of angular momentumI₁ω₁ - I₂ω₂ = 0
I₁ω₁ = I₂ω₂
where;
I₁ is initial moment of inertia I₂ is final moment of inertiaω₁ is initial angular speedω₂ is final angular speedThe final angular speed when the box slides off;
ω₂ = I₁ω₁ / I₂
ω₂ = [0.5(m₁ + m₂)(R + r)²] / (0.5MR²)
ω₂ = [0.5(30 + 0.3)(3.1 + 1.4)²] / (0.5 x 30 x 3.1²)
ω₂ = 2.13 rad/s
Time taken for the for box to slide off;
τ = Iα
τ = I(ω/t)
τ = (Iω)/t
t = (Iω)/τ
t = (0.5 x 0.3 x 1.4² x 2.13)/6
t = 0.1 s
Thus, the time after it starts spinning, that the box will slip off is 0.1 s.
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. population is of great importance in the economic development of Country Explain this statement with four point.
yes
Explanation:
it develops alot of trade
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capacitor take to loose half its charge?
B) How long does it take the capacitor to loose half of its stored energy?
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
Time taken to loose half of its charge
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
Time taken to loose half of its stored energyU(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
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You have two rocks made of the same material that are at the same
temperature. The first rock has twice the mass of the second rock. How does
the thermal energy of the two rocks compare?
A. The second rock has four times the thermal energy of the second
rock.
B. The first rock has half as much thermal energy as the second rock.
OC. The rocks have the same amount of thermal energy.
D. The first rock has twice as much thermal energy as the second
rock.
Answer:
I think its C
Answer:
D
Explanation:
If the temperature doesn't change but the mass of the object increases, the thermal energy in the object increases.
A force F parallel to X-axis is starching a very light horizontal spring. Variation of force F vs. X, elongation of the spring is shown in the figure. (a)What is the spring force constant? (b) Calculate the work done by the force F as the spring elongates from x=4.0 cm to x=8.0 m.
The spring constant is obtained from the ratio of applied force to the elongation of the spring. The work done by the force is 2F.
Spring constant
The spring constant is obtained from the ratio of applied force to the elongation of the spring.
k = F/x (N/m)
Work done by the forceThe work done by the force is calculated as follows;
W = ¹/₂FΔx
where;
F is applied forceΔx is elongation of the spring = 8 m - 4 m = 4 mW = ¹/₂F(4)
W = 2F
Thus, the spring constant is obtained from the ratio of applied force to the elongation of the spring. The work done by the force is 2F.
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An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
The magnitude of the tension in the lower rope is 12.8 N. Calculate the magnitude of the tension in the upper rope.
24.93 N is incorrect.
The tension in the upper rope is determined as 50.53 N.
Tension in the upper rope
The tension in the upper rope is calculated as follows;
T(u) = T(d)+ mg
where;
T(u) is tension in upper ropeT(d) is tension in lower ropeT(u) = 12.8 N + 3.85(9.8)
T(u) = 50.53 N
Thus, the tension in the upper rope is determined as 50.53 N.
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