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Shkiper50 [21]
3 years ago
9

A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later

, the 757 kg car reaches the top of the hill with a speed of 25.0 m/s. How much charge did the batteries move through the motor to accomplish this, neglecting friction

Physics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:

140265.8 C = 1.403 × 10⁵ C

Explanation:

The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.

Potential work required to move the 757 kg car up a vertical height of 195 m = mgh

P.E = 757 × 9.8 × 195 = 1446627 J

Kinetic work done = (1/2)(m)(v²)

K.E = (1/2)(757)(25²) = 236562.5 J

Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J

And this would be equal to the potential of the battery.

For the battery, potential difference = (electric potential energy)/(charges moved)

ΔV = ΔU/q

q = ΔU/ΔV

ΔU = 1683189.5 J

ΔV = 12.0 V

q = 1683189.5/12 = 140265.8 C

Airida [17]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

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A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang
MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0
3 years ago
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (
vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,                      

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg                

angular speed of the disk  = 4 rev/s

arm outstretched = 1 m                  

centripetal force of the disk in the circular path

F = m ω² r                        

ω = 4 x 2 x π        

ω = 25.13 rad/s

F = m ω² r                      

F = 2 x 25.13² x 1

F = 1263.03 N                                              

hence, centripetal force equal to the F = 1263.03 N

6 0
3 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
3 years ago
Read 2 more answers
Open Meet and enter this code: jnw-xodp-yij​
Arada [10]

Answer:

Is this for zoommmmmmmmmm?

Explanation:

It had to be 20 letters I had to do something

5 0
3 years ago
Read 2 more answers
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