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3 years ago

Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?

Physics

1 answer:

Murrr4er [49]3 years ago

8 0

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

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Students are building basic circuits in their physics class and they are asked to measure the current running through their circ

storchak [24]

Answer: option D

Explanation: students will need to make the ammeter a component built into their circuit so that electron flow through it to measure the current, by doing so, they have to connect the ammeter in series with the resistor hence making the same value of current flow through both the ammeter and the resistor.

The value of current on the ammeter is the value of current in the resistor which is the value of current flowing in the circuit as a whole.

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4 years ago

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As a train accelerates away from a station, it reaches a speed of 4.6 m/s in 5.2 s. If the train's acceleration remains constant

Svet_ta [14]

Answer:

Vf = 10.76 m/s

Explanation:

Train kinematics

The train moves with uniformly accelerated movement

$V_f = V_o + a*t$ Formula (1)

Vf: Final speed (m/s)

V₀: Inital speed (m/s)

t: time in seconds (s)

a: acceleration (m/s²)

Movement from t = 0 to t = 5.2s

We replace in formula (1)

4.6 = 0 + a*5.2

a = 4.6/5.2 = 0.88 m/s²

Movement from t = 5.2s to t = 5.2s + 7s = 12.2s

We replace in formula (1)

$V_f = 4.6 + 0.88*7$

Vf = 10.76 m/s

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3 years ago

How do i balance WO3 + H2 = W+ H₂O<br>

Levart [38]

Answer:

Hello, I think it'll be:

WO3 + 3H2 = W + 3H₂O

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3 years ago

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Describe the conversions between potential and kinetic energy that occur when you shoot a basketball at a basket

Aleks04 [339]

When you shoot a basketball at a basket, the energy from your hands is transferred into the basketball. When you hold the ball up in preparation to shoot, it has potential energy, meaning that it has stored energy that can turn into kinetic energy. When you shoot it, that potential energy turns into kinetic energy, causing the basketball to move.

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3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

$\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}$

Now we use the formula

${v_f}^2-{v_i}^2=2a(x_f-x_i)$

to find the distance it travels. The sled comes to a rest, so $v_f=0$ , and let's take the starting position $x_i=0$ to be the origin. Then the distance traveled $x_f-x_i=x_f$ is

$-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m$

6 0

3 years ago