Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
given that
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
now from above equation
So force applied by the person must be 150 N
Answer:the pressure depends on gas and it will be half as much underwater
Explanation:
Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.
This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .
Answer:
The distance covered by the sprinter, s = 40 m
Explanation:
Given data,
The initial velocity of the sprinter, u = 0 m/s
The acceleration of the sprinter, a = 5 m/s²
The time period of acceleration of the sprinter, t = 4 s
Using the II equations of motion
s = ut + ½ at²
= 0 + ½ (5) (4)²
= 40 m
Hence, the distance covered by the sprinter, s = 40 m
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² = 
x² + g x
let's calculate
v² = 
1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s
