Answer:
B. Nuclear Energy
A 0.08m long spring is attached vertically to the ceiling. after 66kg mass is attached to the spring it has a length of 2.1m. determine the spring's force constant. use g=10
Answer:
The spring constant is 647.05 N/m.
Explanation:
The weight of the mass is W=mg
W=66*10
W=660
The spring force is given by the formula F=kx
where k is the spring force constant.
The spring force is equal to the weight of the mass. Therefore equate W and F.
kx= 660
k=660/x
Difference in length x=2.1-0.08= 1.02 m
k=660/1.02
k=647.05 N/m
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How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 joules over grams times degrees celsius) from 21°c to 363°c? the answer is expressed to the nearest whole number. 121 joules 132 joules 140 joules 149 joules
The heat needed is 132J
Given the mass of the gold m = 3 gram.
The initial temperature of the gold T = 21 degree celsius.
The final temperature of the gold T' = 363 degree celsius.
Hence, the change in temperature dT = T' - T
= 363 - 21 degree celsius
= 342 degree celsius.
The specific heat of the gold is given as c = 0.129 J/g°C
We are asked to calculate the heat required ( dQ ) to raise the temperature of gold.
The heat required for this is calculated as
dQ = mcDT
dQ = 3 × 342 × 0.129 J
dQ = 132.354 J ≈ 132 J
Hence, the Heat needed is 132 J.
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Answer: B (132 joules)
Explanation:
6 An infinite non-conducting sheet has a surface charge density o = 5.80pC/m². (a) How
much work is done.
by the
the electric field due to the sheet if a particle of charge
ue to
q+1.60x10-¹°C is moved from the sheet to a point P at distance d= 3.56cm from the
sheets. (b) If the electric potential V is defined to be zero on the sheet, what is V at P
a. Work done = 2. 0648 Joules
b. Electrical potential, V = 4. 03 × 10^10 volts
How to determine the parameters
a. Work done = force × distance
Force = density × acceleration due to gravity
Force = 5. 80 × 10
Force = 58 Newton
We have that distance = 3. 56cm = 0. 0356 m
Work done = 58 × 0. 0356
Work done = 2. 0648 Joules
b. The formula for electric potential is given thus,
Electric potential, V = [tex]k\frac{q}{r}[/tex]
r = 0. 0356m
k = Coulomb's constant = [tex]8.98*10^9[/tex]kg⋅m3⋅s−4⋅A−2.
q= 1.60x10^-¹C
substitute into the formula
Electric potential, V = [tex]8. 98[/tex] × [tex]10^9[/tex] × [tex]\frac{1. 6 * 10^-^-1}{0. 0356}[/tex]
Electric potential, V = [tex]8. 98 * 10^9[/tex] × [tex]4. 494[/tex]
Electric potential, V = [tex]4. 03 * 10^1^0[/tex] volts
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Gneiss and quartzite rocks are found in the metamorphic surface bedrock in which new york state landscape region ?
Answer: It's mostly found near Jamestown , New York
The average velocity of a plane during a flight was 9.0 m/s. How long did it take the plane to travel 5.0 km
Answer:
The time taken is 555.56 s.
Explanation:
The average velocity is the ratio of the net velocity to the time required to achieve net velocity.
The formula of average velocity is,
v(av)=d/t
Given d=5.0 km which is 5*1000 m=5000 m and v(av)=9.0 m/s.
Therefore 9=5000/t
t=5000/9
t=555.56
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A Eureka can of mass 60g and cross sectional area of 60 square centimeters is fillied with water of density 1g/ cubic centimeters. A piece of steel of mass 20g and density 8g/ cubic centimeters is lowered carefully into the can.
(a) Calculate the total mass of water and the Eureka Can before the metal was lowered.
(b) Calculate the volume of the water that overflowed.
(c) Calculate the final mass of the Eureka Can and its contents.
a.600 cm³
b. 2.5 g/cm³
c. 2.0775 kg
What is Eureka moment?
In Physics, this Archimedes' Eureka moment is called the Archimedes Principle, which states that when a body is immersed in a liquid, it experiences an upward buoyant force, which is equal to the weight of the liquid displaced by the body.
According to the question
Mass of Eureka is 60g
Cross sectional area is 60 square centimeters
Mass of steel piece is 20g
Density = 8g /cubic centimeter
a)The total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can.
To find the mass of the water but we can easily find if we know the volume of the can.
In order to calculate the volume we would have to multiply the area of the cross section by the height.
Here,
60cm² x 10cm = 600 cm³
To find the mass that water has in this case we have to multiply the water's density by the volume,
so we get....
[tex]\frac{1g}{cm^3}[/tex] x 1000 = 1000g or 1kg
Now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can.
So we get....
1000g + 60g = 1060g or 1.06kg
b)The volume of the water that over flowed will be equal to the volume of the metal piece since, when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle".
Knowing this we just have to calculate the volume of the metal piece an that will be the answer.
So this time in order to find volume we will have to divide the total mass of the metal piece by its density.
So we get
20g ÷ [tex]\frac{8}{cm^3}[/tex] = 2.5 g/cm³
c)Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece.
We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 overflowed.
So now we the same process as in number a) just with a few adjustments.
[tex]\frac{1g}{cm^3}[/tex] x (1000 - 2.5) = 997.5g
So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can.
The mass of the water, and the mass of the metal piece.
Hence,
1060g + 997.5g + 20g = 2077.5 g or 2.0775 kg
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A tank containing water has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3o angle with the normal to the glass. Find the angle the light makes with the normal in the water.
The normal in the water is 29.8°
Let
[tex]n_i=[/tex] the index of refraction for the material with incident light,
[tex]n _r[/tex] = the index of refraction for the material with refracted light,
[tex]\theta_i=[/tex] the angle of incidence
[tex]\theta_r[/tex] = the angle of refraction.
Applying Snell's law
[tex]n_isin\theta_i=n_rsin\theta_r[/tex]
Given
[tex]n_a=1, n_g=1.550, n_m=1.329n[/tex]
a)
From air to glass
[tex]n_a sin\theta_1[/tex][tex]=n_gsin\theta_2[/tex]
[tex]1\\sin *41.3 = 1.550 *sin\theta_2[/tex]
A normal from glass to water
[tex]n_g sin\theta_2 =n_m sin \theta_3 \\1.550\cdot\sin25.2\degree=1.329\cdot\sin\theta_3\\1.550*sin25.2=1.329*sin\theta_3 \\\theta_3=\arcsin({1.550\cdot \sin25.2\degree \over 1.329})\\\approx29.8\theta_3 =arc sin( 1.3291*550*sin25) ≈ 29.8[/tex]
Hence the normal in the water is 29.8°
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Which is a valid prediction about the continuous function f(x)? f(x) ≤ 0 over the interval (–[infinity], [infinity]). f(x) > 0 over the interval (–1, [infinity]). f(x) ≥ 0 over the interval [–1, 1]. f(x) < 0 over the interval (0, 2).
The valid prediction is f(x) ≥ 0 over the interval [–1, 1]
The question seems to be incomplete the complete question is given in the image!!!!
A function from a set X to a set Y is an assignment of an element of Y to each element of X. The set X is called the domain of the function and the set Y is called the codomain of the function.
The given values for x = 3 , f(x) = -15 Here f (x) < 0
When x =0, f(x) = 5 Here f(x) > 0
So for f(x) ≤ 0 over the interval (–[infinity], [infinity]) is the wrong option
When x = 2 f(x)=-5 here f(x) < 0
So , f(x) > 0 over the interval (–1, [infinity]) is also a wrong option
When x = 1 f(x)=0 here f(x) > 0
So , f(x) < 0 over the interval (0, 2) is also a wrong option
When x = -1 f(x) = 0
When x = 0 f(x) = 5
When x = 1 f(x) =0
So , f(x) ≥ 0 over the interval [–1, 1] is the valid prediction
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When encountering a powerboat in darkness or reduced visibility, what do visible white and red lights indicate?.
A visible white and red light indicate a second craft is coming from the starboard (right) side.
What is a navigation light?A navigation light often referred to as a running light or position light, is a source of illumination aboard a ship, plane, or spacecraft that provides information about the location, course, or condition of the vehicle. Red and green navigation lights help with traffic control by indicating the orientation of the craft.
All navigation light systems typically comprise one or more white lights as well as red and green sidelights that designate the boat's port and starboard sides.
Having a flashlight on board is also essential since you never know when a navigation light may go out.
The typical navigational lights:
SidelightsSternlightMasthead lampoverall white lightingSidelights: Because they are visible to other vessels approaching from the side or head-on, these red and green lights are sometimes known as combo lights. The port (left) and starboard (right) sides of a ship are indicated by the red and green lights, respectively.
sternlight: Only from behind or almost behind the ship can one see the sternlight.
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What are the products of a fusion reaction? Check all that apply.
lighter atoms
energy
heavier atoms
a neutron
a proton
Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.
What are the products of a fusion reaction?
Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.
So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.
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Answer:
B C D
Explanation:
Which equation represents the law of conservation of energy in a closed system? kei pei = kef pef pei pef = kei kef kei – kef = pei – pef kei – pef = pei – kef
In equation form, Or KE+PE=constant KEi+PEi=KEf+PEf where i and f denote initial and final values. This is known as the conservation of mechanical energy.
Κi + Ûg,i = kf + U,g,f
A closed system is a system where exists energy interactions with surroundings, but not mass interactions.
If we neglect any energy interactions from boundary work, heat, electricity, magnetism, and nuclear phenomena and assume that process occurs at a steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduced to this form:
K + Ug = 0 (1)
Where:
K -Change in kinetic energy of the system, measured in joules.
Ug - Change in gravitational potential energy of the system, measured in joules.
If we know that K = Ki - Kf and Ug = Ug,i - Ug,f, then we get the following equation:
Ki + Ug,i = Kf + Ug,f, (2)
Where i and f stand for initial and final states of each energy component.
Hence, the right answer is Ki + Ug,i = Kf + Ug,f,.
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Answer:
A KEi + PEi = KEf + PEf
Explanation:
A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directly above the power source is a dot labeled X and then a circle with an X in it. The circuit then splits with one path straight back to the power source and the other path has 3 circles with X in them labeled 2, 3, and 4 respectively. There is a point labeled Y between circles 2 and 3.A diagram of a circuit with a power source on the left. Directly above the power source is a dot labeled X and then a circle with an X in it. After the x are 4 different circles on the circuit with Xs in them labeled 1, 2, 3, and 4 respectively. There is a point labeled Y in between circles 2 and 3. There is a branch of the circuit from X to Y. How does the circuit change when the wire is added?
A closed circuit occurs and makes all bulbs turn off.
An open circuit occurs and makes all bulbs turn off.
A short circuit occurs and makes bulbs 3 and 4 turn off but keeps bulbs 1 and 2 lit.
A short circuit occurs and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit.
The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.
How does the circuit change when the wire is added?Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.
In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.
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which process receives geo thermal energy naturally
12. A 2-kg soccer ball has a recorded acceleration of 30 m/s² after it is
kicked. What was the force of the kick?
Ο Α.Ο.Ν
OB. 15 N
OC. 30 N
OD. 60 N
1 point
An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 201 m/s2 . What is the direction of the electric field
The force on a charge in an electric field is E = 1*10^-9 N/C.
F = QE
and according to newton's second law, the acceleration the charge will undergo is :
a = F / m
Putting these two equations together we get
E = m/q
For an electron with mass m= 9.11*10^-31 kg and charge Q= 1.60*10^-19 C.
E = 9.11*10^-31 kg / 1.60*10^-19 C
E = 1*10^-9 N/C.
Electric discipline, electric-powered assets associated with each point in space when the price is present in any form. The magnitude and direction of the electrical subject are expressed with the aid of the value of E, called electric area power or electric powered discipline intensity or certainly the electrical area.
An electric-powered discipline is a place of the area around an electrically charged particle or objects wherein an electric charge might feel pressure. An electric-powered field is a vector amount and can be visualized as arrows going towards or far from expenses.
Examples of electric fields consist of the field produced in the dielectric of a parallel-plate capacitor (which creates an electrostatic discipline) and the electromagnetic wave produced by a radio broadcast monopole antenna (which creates a time-various area)
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Along the frictionless path you have chosen in question 8, the main force(s) acting on the puck after receiving the kick is (are):
O a downward force of gravity.
a downward force of gravity, and a horizontal force in the direction of motion.
a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motion.
a downward force of gravity and an upward force exerted by the surface.
none. (No forces act on the puck.)
The main force(s) acting on the puck after receiving the kick is a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motion.
What is friction?Friction can be defined as that resistance that one surface or object encounters when moving over another.
So therefore, the main force(s) acting on the puck after receiving the kick is a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motion
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If the current through a resistor is tripled, how does the power dissipated by
the resistor change?
If the current that is circuit is tripled, the power is increased by a factor of 9. Option D
What is power?The term power refers to the rate of doing work. Now if we have to obtain the work done by the use of the formula; P = I^2R
Now if the current that is circuit is tripled, the power is now obtained by;
P = (3I)^2R = 9I^2R hence the power is increased by a factor of 9.
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what is magnetic shielding?
Spiderman, whose mass is 80.0 kg, is dangling on the free end of a 12.0-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a 60.08 angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver
4.719kJ is the work done by the gravitational force.
Given,
Mass of Spiderman, m = 80 kg
Acceleration due to gravity, g = 9.8[tex]m/s^{2}[/tex]
Formula used
F= mg
So, The force on Spiderman is= 80 × 9.8 = 784N
As Spiderman is dangling on the free end of a long rope, the other end of which is fixed to a tree limb above.
That's why only his vertical movement makes a difference in how much work gravity does.
So, we only consider the y coordinates here
As the rope is 12m long ,
y coordinate below the tree limb is –12m. (negative sign because the value is present below the origin.)
As it is given that the rope makes a 60.08° angle with the vertical.
So his final y coordinate is (–12.0m)cos60.08°
= -12×0.498 = -5.98m
Change in elevation (ΔS)= –5.98m–(–12.0m)= 6.02m.
The work done by the gravity on Spiderman is:
W=FΔscosθ….. (1)
where, F = Force of gravity on Spiderman
ΔS = Change in the elevation
θ = angle between force and the displacement
Putting the values in equation (1)
W = 784 × 6.02 ×cos 180°
= 784 × 6.02 × -1 = 4719.68 J = 4.719kJ
Hence, the work was done by the gravitational force on Spiderman in this maneuver is 4.719 kJ.
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A certain motor is capable of doing 5600 joules of work in 10 seconds. What is the power output of this motor?
Answer:
Power output of the motor is 560W.Explanation:
Power output = ?Work done = 5600JTime = 10sPower output = work done/timePower output = 5600J/10sPower output = 560WTherefore the Power output of the motor is 560W.Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev>min?
3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
[tex]v=rw[/tex] where,
[tex]v[/tex] is the Linear velocity
[tex]r[/tex] is the radius
[tex]w[/tex] is the angular velocity
[tex]v[/tex] = 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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YALL PLEASE HELP QUICK!!!!!!!11!!!1!!1
4.16 Lab Report
Motion and Forces
Introduction
1. What was the purpose of the experiment?
2. What were the independent, dependent, and control variables in your investigation? Describe the variables for each part of the experiment.
3. Write a hypothesis based on observations and scientific principles.
Experimental Methods
1. What tools did you use to collect your data?
2. Write your procedure. List each step so that another student could follow the procedure and repeat your experiment.
Data and Observations
1. Record your observations.
Conclusions
1. What conclusions can you draw about how the sum of forces acting on an object affect its motion? Write an evidence-based claim.
2. Which mass, A, B, or C, was the greatest? Use the observations from your experiment and your knowledge of Newton’s laws of motion to write an evidence-based claim.
The purpose of the experiment is to describe how the plasma membrane contributes to how the cell functions.
How to explain the experiment?The dependent variable is the variable bring investigated. In this case, the dependent variable is the egg circumference.
The independent variable is the variable controlled by the researcher.
The controlled variable has no influence on the results. An example is the concentration of vinegar.
A hypothesis is a proposition which can be tested by observations. In order to form a hypothesis, it's important to collect observations about the problem being examined.
The tools that can be used to collect data include observations and records taken.
The procedure to collect the data include:
Determine the information be collected.Set a timeline for the collection of data.Determine the method to utilize for data collection.Analyze the data.Implement the findings.The conclusions that can be drawn about how the sum of forces acting on an object affect its motion is that when the forces cancel out, then the resultant force will be zero, if not, the unbalanced force will result in the acceleration of the object.
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April is moving into a new apartment. She is moving a box of books that has a mass of 30 kg.
a. What is the weight of the box? (2 points)
b. What is the normal force acting on the box when it sits on the floor? Explain how you found the answer. (2 points)
c. If she pushes on the box with a force of 100 N and there is no friction, what is the acceleration of the box? (2 points)
d. If the coefficient of static friction between the box and the floor is 0.3, what is the maximum force of static friction acting on the box? (2 points)
e. How hard must she push on the box to get it to slide? (2 points)
f. If she pushes on the box with a force of 100 N, and the coefficient of kinetic friction is 0.25, what is the net force on the box in the x-direction? (2 points)
g. Based on your answer to part (f), what is the acceleration of the box? (2 points)
The weight of the box is 300 newton whereas the acceleration is 3.3 m/s2.
What is the weight of the box?The weight of the box is 300 newton because weight is a force which move towards the center of the earth due to gravity and we know that gravity is 10 m/s2.
What is the acceleration of the box?The box is moving with 3.3 m/s2 if we put in the formula below.
acceleration= force/mass
A = 100/30
A = 3.3 m/s2
So we can conclude that the weight of the box is 300 newton whereas the acceleration is 3.3 m/s2.
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A rock in space travels in the x direction with a speed of 6 m/s and collides with an identical rock at rest. After the collision, the first rock is traveling at 4 m/s at an angle of 60 degrees with respect to the x axis. What is the speed of the second rock after the collision, in m/s
The speed of the second rock after the collision is 4.15m/s
Given the direction of rock in space is x , the direction with speed is 6m/s
After the collision the speed of the first rock is at 4 m/s
The angle given is 60 degrees with respect to x - axis
We need to calculate the speed of the second rock after the collision is
We know that
[tex]m_1=m_2[/tex]
[tex]v_1 = 6m/s\\v_2 = 4m/s[/tex]
θ = 60°
We need to find [tex]v_x[/tex] and [tex]v_y[/tex]
Therefore ,
[tex]m_1v_1= m_2v_2cos60+m_2v_2\\[/tex]
So, 6 = 4 cos 60° + [tex]v_2[/tex]
[tex]v_x = 6-4cos60\\v_x = 2m/s[/tex]
The same way we will find the momentum for [tex]v_y[/tex]
The total momentum is 0
0+0 = [tex]m_1(4)sin60 + m_2v_y[/tex]
[tex]v_y = -3.64 m/s[/tex]
Now ,
V (resultant) = [tex]\sqrt{(v_x)^{2} + (v_y)^2} \\v = \sqrt{(2)^2+(-3.64)^2} \\v = 4.15 m/s[/tex]
Hence the speed of the second rock after the collision, in m/s is 4.15m/s
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1. You released a pendulum of mass 1kg from a height of 0.05m
b. If the pendulum loses 18% of the initial energy by the time it reaches the bottom, how fast is it going when it reaches the bottom?
c. If the pendulum loses another 7% of its remaining energy by the time it reaches the other side, how high up does it go?
d. After a few minutes the pendulum is no longer swinging at all, explain why this happens in terms of energy?
a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.
b. The height reached by the pendulum is 0.038 m.
c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.
Kinetic energy of the pendulum when it reaches bottomK.E = 100%P.E - 18%P.E
where;
P.E is potential; energyK.E(bottom) = 0.82P.E
K.E(bottom) = 0.82(mgh)
K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J
Speed of the pendulumK.E = ¹/₂mv²
2K.E = mv²
v² = (2K.E)/m
v² = (2 x 0.402)/1
v² = 0.804
v = √0.804
v = 0.9 m/s
Final potential energyP.E = 100%K.E - 7%K.E
P.E = 93%K.E
P.E = 0.93(0.402 J)
P.E = 0.374 J
Height reached by the pendulumP.E = mgh
h = P.E/mg
h = (0.374)/(1 x 9.8)
h = 0.038 m
when the pendulum stopsWhen the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.
Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.
The height reached by the pendulum is 0.038 m.
When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.
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A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displacement is 1. 75 cm?.
0.22 m/s is the speed when the displacement is 1.75 cm.
Given:
A simple harmonic oscillator has;
Amplitude (A) =3.50 cm = 0.035 m
Maximum Speed (Vmax) = 26.0 cm/s = 0.26 m/s
Displacment (d) = 1.75cm =0.0175 m
The displacement d, whose maximum is the amplitude A , is expressed as:
∴ d = A Sin wt
[tex]\frac{d}{A}[/tex] = Sin wt
t = [tex]\frac{1}{W}[/tex] Sin⁻¹ ( [tex]\frac{d}{A}[/tex] )
v = - Aw cos wt
v = - Aw cos w [ [tex]\frac{1}{W}[/tex]sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]
v = - Aw cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ]
Speed, v = Vmax cos [ sin⁻¹ ([tex]\frac{d}{A}[/tex]) ] ∵ Vmax = Aw
v = 0.26 cos [ sin⁻¹ ( [tex]\frac{0.0175}{0.035}[/tex]) ]
v = 0.22 m/s
Therefore, 0.22 m/s is the speed when the displacement is 1.75 cm.
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A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. The same car under similar road conditions, traveling at 140 m/s, takes ______________ m to stop under full braking
A [tex]1000[/tex] kg car travelling with velocity [tex]70[/tex]m/s takes [tex]3[/tex] to stop under full braking. And the same car under similar road conditions travelling with velocity [tex]140m/s[/tex] takes [tex]12m[/tex] to stop under full braking.
How can we find Distance at velocity [tex]140m/s[/tex] ?
In case of [tex]70m/s[/tex] car stop under full braking after [tex]3m[/tex].
So we can find the acceleration by third equation of motion
[tex]v^2=u^2+2as[/tex]
Here [tex]v[/tex] is final velocity which is zero.
[tex]u[/tex] is initial velocity which is [tex]70m/s[/tex]
[tex]s[/tex] is displacement
So substitute the values in equation of motion
[tex]0=70^2+2*a*3\\-4900/6=a\\-816.6m/s^2=a[/tex]
In case of initial velocity [tex]140m/s[/tex] acceleration is same because same car and same road condition so acceleration is [tex]-816.6m/s^2[/tex]
We can find the distance by using the equation of motion
[tex]v^2=u^2+2as\\0=140^2+2*(-816.6)*s\\-19600/-1633.2=s\\12m=s[/tex]
Distance is [tex]12m[/tex]
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You took a second ball headed to a cliff. You dropped this ball straight down with a velocity of 9.3 m/s. The ball hit the ground exactly 8.0s after you released it. How high is the ciff?
Answer:
The cliff is 74.4m.
Explanation:
The reading above clearly state that a ball is dropped straight down with the speed of 9.3m/s and hit the ground exactly at a time of 8.0s after you released it. And it is asking for the distance of the cliff.
In speed formula make distance the subject.
Therefore Distance = speed × time
Distance = 9.3mls × 8.0s = 74.4m
Therefore the cliff is 74.4m.
HELP PLEASE!!
identify health-related problems related to an unhealthy body composition.
( I didn't know what subject to put this as, sorry )
Some health-related problems that are related to unhealthy body composition are:
Type 2 DiabetesHigh blood pressureStrokeApneaWhat is a Health Problem?This refers to the complications that a person has when his health system does not function optimally.
Hence, we can see that based on the fact that the body needs to have a healthy body composition such as a good Body Mass Index when this becomes unbalanced, it can bring the above health problems.
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Need help in knowing what energy it’s using
Answer:
8:P.E,9:gaining K.E,10:P.E,11:gaining K.E.
Explanation:
it gained potential energy at 8 because it was at its maximum height,and since potential energy is the energy that u posses due to ur height,thats the energy it possesed on the spot 8.as you can see on spot 9 its moving down so as its moving down,it gains kinetic energy,since its an energy possesed by a body due to its motion/movement,thats what it gained at spot 9, it gsained kineti energy because of its movement.and at 10 agian it would gain potential energy ecause of height and on point 11 as its heading upwards it gains kinetic energy because of movement.