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rosijanka [135]
3 years ago
14

How do i solve?a = bc(b)​

Physics
1 answer:
stira [4]3 years ago
7 0
I don’t even know I’m so dumb.
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Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0
4 years ago
35 POINTS! Say If You Drop A Ball from 100 Centimeters. When The ball Bounces, The Ball Does Not Bounce Back Up To 100 Centimete
diamong [38]

some ball when you bounce it it comes back up but according to gravity the energy goes away

8 0
3 years ago
Which graph shows the correct relationship between kinetic and speed
Elenna [48]
I can’t answer without any graph options
6 0
3 years ago
Read 2 more answers
The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. an ostrich is run
Orlov [11]
Velocity of an object is its rate of change of the object's position per interval of time. Velocity is a vector quantity which means that it consists of a magnitude and a direction. Magnitude is represented by the speed and the direction is represented by the angle. To determine the velocity components, we use trigonometric functions to determine the angle of the components. For the north component we, use the sine function while, for the west component, we use the cosine function. We calculate as follows:
north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
<span>west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s</span>



6 0
4 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le
xz_007 [3.2K]

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

4 0
3 years ago
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