Answer:
<u><em>0.03 m/s</em></u>
Explanation:
<em>Applying law of conservation of momentum, </em>
- <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
- <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
- <em>75.105v = 2.52</em>
- <em>v = 2.52/75.105</em>
- <em>v = </em><u><em>0.03 m/s</em></u>
Newtons first law of motion is also known as the law of inertia
Answer:
40 m/s
Explanation:
given,
height of the fall, h = 82 m
time taken to fall, t = 1.3 s
rock velocity, v = ?
acceleration due to gravity, g = 9.8 m/s²
rock is released initial velocity, u = 0 m/s
using equation of motion
v² = u² + 2 a s
v² = 0 + 2 x 9.8 x 82
v² = 1607.2
v = 40 m/s
hence, rock's velocity is equal to 40 m/s
Answer:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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