PE = (mass) (gravity) (height)
PE = (0.005 kg) (9.8 m/s²) (5 m)
<em>PE = 0.245 Joule</em>
Answer:
Approximately 
, assuming that the acceleration of this ball is constant during the descent.
Explanation:
Assume that the acceleration of this ball, 
, is constant during the entire descent.
Let 
denote the displacement of this ball and let 
denote the duration of the descent. The SUVAT equation 
would apply.
Rearrange this equation to find an expression for the acceleration, , of this ball:
.
Note that 
and 
in this question. Thus:
.
Let 
denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is , the net external force on this ball would be 
.
Since 
and 
, the net external force on this ball would be:
.
Answer:
the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:
This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is
fr- w || = 0
The expression for friction force is that it is proportional to the coefficient of friction by normal.
fr = μ N
When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.
In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.
μ kinetic <μ static
In all this movement the normal with changed that the angle of the table remains fixed.
Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction
Answer:
Explanation:
Let pressure at surface of earth be P Pa.
pressure at height of 8.1 km in air can be calculated as follows .
pressure due to column of air of 8.1 km height
= h d g , h is height , d is density of air and g is acceleration due to gravity
= 8.1 x 1000 x .87 x 9.8 = 6.9 x 10⁴ Pa .
pressure at the height of 8.1 km
= P - 6.9 x 10⁴ Pa
Pressure due to column of 16 m in the sea
= h d g
16 x 1000 x 9.8
= 15.68 x 10⁴ Pa .
Pressure at depth of 16m
= P + 15.68 x 10⁴
pressure difference between points at height of 8.1 km and pressure at point 16 m deep
= P + 15.68 x 10⁴ - P + 6.9 x 10⁴ Pa
= 22.58 x 10⁴ Pa .
