How many sig figs are in: 103 cm ?

A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago

A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

shepuryov [24]

Explanation:

Hello there!!!

You just need to use simple formula for force and momentum, 

F= m.a

and momentum (p)= m.v

where m= mass

v= velocity.

a= acceleration .

And the solutions are in pictures. 

Hope it helps..

5 0

3 years ago

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h

Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft$

The minimum stopping distance is 141.78 ft.

4 0

2 years ago

A 10kg sphere hits a stationary 8kg sphere. After the collision the 8kg sphere moves off in the positive direction at 4m/s. If t

Bumek [7]

M = mass of the first sphere = 10 kg

m = mass of the second sphere = 8 kg

V = initial velocity of the first sphere before collision = 10 m/s

v = initial velocity of the second sphere before collision = 0 m/s

V' = final velocity of the first sphere after collision = ?

v' = final velocity of the second sphere after collision = 4 m/s

using conservation of momentum

M V + m v = M V' + m v'

(10) (10) + (8) (0) = (10) V' + (8) (4)

100 = (10) V' + 32

(10) V' = 68

V' = 6.8 m/s

7 0

2 years ago

Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angul

vredina [299]

Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

L₀ = I w₀

Final

$L_{f}$ = $I_{f}$ $w_{f}$

L₀ = $L_{f}$

I w₀ = $I_{f}$ $w_{f}$

$w_{f}$ = I / $I_{f}$ w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

5 0

2 years ago