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Phoenix [80]
4 years ago
13

A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?

Physics
2 answers:
9966 [12]4 years ago
8 0

The answer would be A

ipn [44]4 years ago
3 0
Your answer would be A. You divide 96 by 16 to find the answer
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A ball whose mass is 0.4 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c
e-lub [12.9K]

Answer:

Explanation:

We shall apply concept of impulse

Impulse = Force x time

= Force x 2 x 10⁻³ N.s

impulse = change in momentum

change in momentum

= .4 x 4 - ( - .4 x 2 )

= 2.4

Force x 2 x 10⁻³  = 2.4

Force = 2.4 /  2 x 10⁻³

= 1.2 x 10³ N .

average magnitude of the force exerted by floor = 1.2 x 10³ N

If R be reaction force by earth

R - mg = 1.2 x 10³

R = 1.2 x 10³ + mg

= 1200 + .4 x 9.8

= 1200 +3.92

= 1203 .92 N .

7 0
3 years ago
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
3 years ago
Help.me on question 3 pls as fast as you can
antoniya [11.8K]

Answer:

Drought

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7 0
3 years ago
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

6 0
3 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

brainly.com/question/14384780

#SPJ1

5 0
1 year ago
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