Given:
m(mass of the box)=10 Kg
t(time of impact)=4 sec
u(initial velocity)=0.(as the body is initially at rest).
v(final velocity)=25m/s
Now we know that
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration acting on the body
t is the time of impact
Substituting these values we get
25=0+a x 4
4a=25
a=6.25m/s^2
Now we also know that
F=mxa
F=10 x6.25
F=62.5N
Answer:
Natalie says that all things with mass have a gravitational field, but the force is very weak and cannot be perceived around small objects.
Explanation:
The force due to gravity is proportional to the mass of the object and inversely proportional to the square of the distance between objects. The Earth is so massive that the force due to its gravity is much greater than the force between objects on the counter.
If there were no friction, the objects might move toward each other, depending on what other masses were near them tending to cause them to move in other directions.
Natalie's explanation is about the best.
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<em>Additional comment</em>
The universal gravitational constant was determined by Henry Cavendish in the late 18th century using lead balls weighing 1.6 pounds and 348 pounds. His experiment was enclosed in a large wooden box to minimize outside effects. While these masses are somewhat greater than those of a glue bottle and stapler, the experiment shows the force of gravity between "small" objects <em>can</em> be measured.
As long as they're both on the same planet, the greater mass always has the greater weight. In this question, Object-A has the greater mass, so it weighs more that Object-B does.
Newton's 2nd law of motion:
Net Force = (mass) x (acceleration) .
The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.
If you know any two of the quantities in the formula,
the law can be used to calculate the third one.
Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.