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hammer [34]
3 years ago
11

Two identical balls collide one was initially at rest. after the collision one of the balls has a velocity of 3 m/s south. the o

ther has a velocity of 4 m/s west. what was the velocity of the ball that was moving before the collision
Physics
1 answer:
Butoxors [25]3 years ago
4 0
Two identical balls collide<span> head on. The </span>initial velocity<span> of </span>one<span> is 0.75 </span>m/s<span> east, while that of the </span>other one<span> is 0.43 </span>m/s west<span>.</span>
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A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
Define escape velocity ​
lianna [129]

Answer:

the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.

Explanation:

8 0
2 years ago
Read 2 more answers
I NEED HELP ASAP PLEASE!!!!!
Gre4nikov [31]

Answer:

answer is 2 option because more force is applied

6 0
2 years ago
1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve
hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

d_x = 12 km east

d_y = 16 km south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

S=\frac{1}{2}gt^2

From which we find the total time of the fall, t:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

d=vt=(16 m/s)(2 s)=32 m

So, the closest answer is B.

5 0
3 years ago
A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t
maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and \theta is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

y=vt+\frac{1}{2}at^2

where y=vertical\ distance

a=acceleration

t=time

v=initial\ velocity

here initial velocity is v\sin \theta [/tex] so

y=v\sin \theta \times 1-\frac{1}{2}gt^2

y=v\sin \theta -0.5g

In absence of gravity

y_2=v\sin \theta \times t

y_2=v\sin \theta \times 1

\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m

         

4 0
3 years ago
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