What are the two factors that determine an objects gravitational potential energy

A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push bo

garik1379 [7]

Answer:

The same

Explanation:

The change in momentum of each cart is equal to the impulse given by the force applied on the cart:

$\Delta p = I = F \Delta t$

where

F is the force applied

$\Delta t$ is the time during which the force is applied

The force applied to both carts (F) is the same, as well as the time ( $\Delta t$ ). This means that the change in momentum, $\Delta p$ , is the same for both carts. But both carts start from rest (momentum = 0): this means that at the end, both carts will have exactly same momentum.

8 0

3 years ago

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0

3 years ago

A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

garik1379 [7]

U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2

6 0

3 years ago

Read 2 more answers

A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on i

Maurinko [17]

Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

5 0

2 years ago

A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0

3 years ago