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yan [13]
4 years ago
5

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a

frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. 1)Which box ends up moving faster?
Physics
1 answer:
11111nata11111 [884]4 years ago
6 0

Answer:

The box 1 moves faster.

Explanation:

lets

Mass =m  kg

Initial velocity = u m/s

Initial velocity of box = 0 m/s

Let stake mass of block = m

When ball bounces back:

The final speed of the box = v

Final speed of ball = - u

Pi = Pf  ( From linear momentum conservation)

m x u + m x 0 = m ( - u) + m v

mu + mu = m v

v= 2 u

When ball get stuck :

The final speed of ball and box = v

Pi = Pf  ( From linear momentum conservation)

m x u + m x 0 = (m+m) v

v= u /2

So the box 1 moves faster.

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mass is 5 lb attached to a rope wound around a pulley. The radius of the pulley is 3 in. If the mass falls at a constant velocit
suter [353]

Answer:

The power transmitted to the pulley is 0.0455 hp.

Explanation:

Given;

mass attached to the rope, m = 5 lb

radius of the pulley, r = 3 in

constant rate of fall of the mass, v = 5 ft/s

acceleration due to gravity, g = 32.2 ft/s²

1 lbf = 32.2 lb.ft/s²

The power transmitted to the pulley is calculated as;

P = Fv

P = (mg)v

P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2}  \ \ = 25 \ \frac{ft.lbf}{s} \\\\

in horse power, the power transmitted is calculated as;

P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s}  \ \ = 0.0455 \ hp

Therefore, the power transmitted to the pulley is 0.0455 hp.

3 0
3 years ago
At takeoff, the horizontal and vertical velocity of long jumper are 57.8 m/s and 5.6 m/s respectively. What is the resultant vel
Lady bird [3.3K]

Apply the Pythagorean theorem to get the resultant velocity:

V = \sqrt{Vx^{2}+Vy^{2}}

Given values:

Vx = 57.8m/s

Vy = 5.6m/s

Plug in and solve for V:

V = 58.1m/s

EDIT: Let's get the direction of the resultant velocity as well.

This equation will give the angle of the velocity as measured off of the ground:

θ = tan⁻¹(Vy/Vx)

Again, the given values are:

Vx = 57.8m/s

Vy = 5.6m/s

Plug in the values and solve for the angle θ:

θ = tan⁻¹(5.6/57.8)

θ = 5.5°

The resultant velocity is oriented 5.5° off the ground.

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How long would it take for someone to bike to school if the school is 10 km away and the
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Answer:

It would take the cyclist 2 hours to cycle to school.

Explanation:

5km/hr for 2 hours would be 10km.

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Speed = 11000 m/s = 11km/s

D = 380000 km,

t = D/s = 380000 km/ 11km/s

t = 34 545.45 seconds.
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