All categories

3 years ago

The work done on a box is 532 joules. The force applied to the box was 48 N. What was the displacement of the box? *

Physics

1 answer:

Semenov [28]3 years ago

8 0

Explanation:

Work = force × displacement

532 J = 48 N × d

d ≈ 11 m

You might be interested in

What is the minimum tangential velocity a space station would need to simulate earth gravity if the radius is 50 meters ?

Aneli [31]

Explanation:

angular velocity is given by

$w = \sqrt{ \frac{g}{r} }$

$w = \sqrt{ \frac{9.8}{25} }$

w = 0.626

now tangential velocity is

V = rw

= 25 x 0.626

= 15.65 m/s

5 0

3 years ago

Describe the velocity of the object shown by the graph .

mrs_skeptik [129]

Answer:

parallel lines

Explanation:

because the line is going down

8 0

3 years ago

Read 2 more answers

Objects a and b each have a mass of 25 kilograms. object a has a velocity of 5.98 meters/second. object b is stationary. they un

mash [69]

The total kinetic energy of the system after collision is 223.5J

In elastic collision kinetic energy and momentum are conserved.

According to the question

mass of object a = 25kg

mass of object b = 25 kg

initial velocity of a u1 = 5.98 m/s

initial velocity of b u2 = 0

so from momentum conservation-

mau1 + mbu2 = (m1+m2)v

25kg × 5.98m/s + 25×0 = (25+25)v

v = 2.99 m/s

Now the total kinetic energy after the collision will be:

final kinetic energy = 1/2 (m1+m2) v²

= 1/2 (25+25)× (2.99)²

= 223.5 J

Thus, total kinetic energy of the system after collision is 223.5J

Learn more about elastic collision here:

brainly.com/question/1808045

#SPJ4

7 0

2 years ago

On a sunny day, your clothing becomes hot because

Brilliant_brown [7]

Answer:

It is A.Heat is transmitted and light is reflected.

4 0

3 years ago

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/

never [62]

Answer:

A. 2.83 m/s

B. 39.55 degrees north of east

Explanation:

The velocity of the ball relative to the ground is the sum vectors of velocity of the ball relative to Mia and the velocity of Mia relative to the ground

If we take north east as positive direction then

Velocity vector of the ball relative to Mia is

$= =$

Velocity of Mia relative to gorund is

<0, 5.3>

So velocity of the ball relative to the ground is

<1.8, -3.12> + <0, 5.3> = <1.8, 2.18>

Its magnitude is:

$v = \sqrt{v_e^2 + v_n^2} = \sqrt{1.8^2 + 2.18^2} = \sqrt{3.24 + 4.7524} = \sqrt{7.9924} = 2.83$ m/s

Its direction is

$tan\alpha = \frac{v_e}{v_n} = \frac{1.8}{2.18} = 0.83$

$\alpha = tan^{-1}0.83 = 0.69 rad \approx 39.55 degrees$ north of east

4 0

3 years ago