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dolphi86 [110]
3 years ago
15

WORK + ENERGY: CONCEPTUAL QUESTIONS. 15) A satellite is in a circular orbit above Earth's surface. Why is the work done on the s

atellite by the gravitational force zero? What does the work-kinetic energy theorem predict about the satellite's speed?. 16) A car traveling at 50.0 km/h skids a distance of 35 m after its brakes lock. Estimate how far it will skid if its brakes lock when itsinital speed is 100.0 km/h. What happens to the car's kinetic energy as it comes to rest?. 17) Explain why more energy is needed to walk up stairs than to walk horizontally at the same speed. 18) How can the work-kinetic energy theorem explain why the force of sliding friction reduces the kinetic energy of a particle?
Physics
2 answers:
Vinil7 [7]3 years ago
4 0
For number 15, the work that was done on the satellite force is zero because in space, there are no other forces or work that is acting upon an object. Based on the work-kinetic energy theorem, there was no change in speed of because there was not work done and change in kinetic energy.

For number 16, 
W = d (K.E.)
35/d = (50)²/(100)²
35/d= 1/4
d= 35 x 4
d= 140 m

It will skid for 140 meters. Based on the law of the conservation of energy, kinetic energy is converted into sound, heat, and motion once the car goes to a complete stop.

For number 17, the energy required to walk up the stairs is higher compared to walking in a horizontal plane at the same speed because walking up the stairs has both vertical and horizontal components which adds up more distance. Walking up the stairs is doing work against gravity.

For number 18, Wfnet = delta KE
The equation of work energy theorem states that particle speed becomes lowers as the force of sliding friction is the only force that is acting on the object.
Alex17521 [72]3 years ago
4 0
The correct answers to these questions are:
15. Work = 0. The reason why is because in space, there are no other forces or work that is acting upon an object. 
16. <span>d = 140 m</span>
W = d (K.E.)
35/d = (50)^2 / (100)^2
35/d = 1/4
d= 35 * 4
d= 140 m
17. There is more energy used in walking up the stairs than walking in a horizontal plane given the same speed. Walking up the stairs is doing work against gravity.
18. Wfnet = delta KE
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A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0
2 years ago
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Si la fuerza de repulsión entre dos cargas es 18 × 1013
sattari [20]

Answer:

Explanation:

F = kQq/r²

r = √(kQq/F)

a)  r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m

  r = 40 mm

b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m

   r = 42 mm

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A mouse ran 100 centimeters in
olga_2 [115]

Answer:

his speed is 5cm a second

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A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i
Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

m_1v_1+m_2v_2=(m_1+m_2)V

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

v_2=-\dfrac{m_1v_1}{m_2}

v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}

v_2=-1.47\ m/s

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

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3 years ago
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