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3 years ago

A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and

Physics

1 answer:

Kryger [21]3 years ago

8 0

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

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John is running down the street and hears dogs barking in the distance. How do the sound waves change as John approaches the bar

FrozenT [24]

Answer:

The height of the sound waves increases

Explanation:

3 0

4 years ago

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

4 years ago

a concrete cube of side 0.50 m and uniform density 2.0 x 103 kg m–3 is lifted 3.0 m vertically by a crane. what is the change in

AlladinOne [14]

Answer:

Change in potential energy = 7350 Joules

Explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube, $d=2\times 10^3\ kg/m^3$

The cube is lifted vertically by a crane to a height of 3 m

We know that, density $d=\dfrac{m}{V}$

So, m = d × V (V = volume of cube = a³)

$m=2\times 10^3\ kg/m^3\times (0.5\ m)^3$

m = 250 kg

We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.

Potential energy at height h is given by :

$PE=mgh$

PE = 250 kg × 9.8 m/s² ×3 m

PE = 7350 Joules

So, change in potential energy of the cube is 7350 Joules.

5 0

3 years ago

Read 2 more answers

The process of wind blowing sand from one location to another is called

lesya [120]

Answer:

weathering

Explanation:

3 0

3 years ago

Read 2 more answers

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago