Ask question

Ask question

All categories

love history [14]

3 years ago

13

. You have two carts, one which is empty and has mass m. The second cart is of the same mass but loaded with twice the mass of t

he empty cart i.e. it has mass 3m. You push each of them (one at a time) with the same constant force, over the same distance, starting from rest. After you have pushed them through this distance, you remove the force. How will the kinetic energy of the loaded and empty carts compare to each other

1 answer:

frez [133]3 years ago

5 0

Answer:

Their kinetic energies would be the same

Explanation:

This is because, since the force, F acting on them moves the same distance, d, the work done by the force is W = Fd.

Now, from work-kinetic energy principles,

W = ΔK where ΔK = change in kinetic energy of the carts.

Since the work-done is the same for both carts, their change in kinetic energies would also be the same.

Since they start from rest, ΔK = K' - K = K' - 0 = K'

So, the kinetic energies of the carts would be the same

You might be interested in

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

$R=\sqrt{A^2 +B^2 +2AB cos \theta}$

where

A is the magnitude of the first vector

B is the magnitude of the second vector

$\theta$ is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function $cos \theta$ . The value of $cos \theta$ are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

$R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B$

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

$R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}$

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

$R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B$

4 0

4 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is

gavmur [86]

Kelvin is a base unit of temperature scale from SI that defines as zero degree Kelvin (absolute zero). The absolute zero is a hypothetical statement that all molecular movement stops because there is no transient of energy for the molecules to move. When converting temperature in degree Celsius to Kelvin, add 273. You are given 600K and you are asked to find it in degrees Celsius.

T(K) = T(C) + 273
600 K = T(C) + 273
T(C) = 600 – 273
T(C) = 327 °C
<span>The answer is letter B.</span>

6 0

3 years ago

Read 2 more answers

What does a Lewis structure diagram represent?

densk [106]

Answer:

correct me if i'm wrong nut i thinks its a The atomic symbol of an element surrounded by valence electrons

Explanation:

5 0

3 years ago

Read 2 more answers

How do very small objects behave?

nataly862011 [7]

A. very small objects behave like like particles.

7 0

4 years ago

Read 2 more answers