Answer:
Force, 
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

Finally, it stops, v = 0



F = -22713.92 N

So, the magnitude of the force that stops the bullet is 
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
Answer:
Height above a surface
Explanation:
Gravitational potential energy is the energy which an object possesses due to its position above a surface.
It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.
It is given mathematically as:
P. E. = m*g*h
where m = mass of the body
g = acceleration due to gravity
h = height above a surface
m*g represents the weight of the object.
Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.
Answer:
Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading
E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number
Explanation:
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