A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block
2 answers:
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(a) At level A Potential Energy is 2940 J.
(b) At level B Potential Energy is 1960 J.
(c) At level C Potential energy is 1470 J.
(d) At level D Potential energy is 0 J.
(e) The change in Potential energy if the object moves from A to B is 980 J.
<u>Explanation</u>:
Given that,
Mass of an object is 5 kg (m) at a height of 60 m (h) above the ground as shown in the given figure.
Potential energy of an object is calculated by using the formula m × g × h. “g” is acceleration due to gravity on earth is 9.8 .
(a) At level A:
At height (h) is 60 m above ground.
Potential energy = 5 × 9.8 × 60
Potential energy = 2940 J
(b) At level B:
At height (h) is 40 m above ground.
Potential energy = 5 × 9.8 × 40
Potential energy = 1960 J
(c) At level C:
At height (h) is 30 m above ground.
Potential energy = 5 × 9.8 × 30
Potential energy = 1470 J
(d) At level D:
At height (h) is 0 m above ground (object is on the ground).
Potential energy = 5 × 9.8 × 0
Potential energy = 0 J
(e) Change in “Potential energy” when it moves from “A” to “B”
Change in “Potential energy” in difference in initial “Potential energy” (at level A) and final “Potential energy” (at level B)
= 2940 - 1960
= 980 J
Therefore, change in “Potential energy” is 980 J.
Answer:
300000.01008 Pa
123.76237 m/s²
Explanation:
= Density of liquid nitrogen = 808 kg/m³
h = Depth
g = Acceleration due to gravity
P = Atmospheric pressure
Absolute Pressure is given by
Below 2 m from surface
Below 5 m from surface
Subtracting the above equations we get
The acceleration due to gravity on the planet is 123.76237 m/s²
Equating the value of g in the first equation
The atmospheric pressure on the planet is 300000.01008 Pa
Answer:
The moment of inertia of large ring is 2MR².
(A) is correct option.
Explanation:
Given that,
Mass of ring = M
Radius of ring = R
Moment of inertia of a thin ring = MR²
Moment of inertia :
Moment of inertia is the product of the mass of the ring and square of radius of the ring.
We need to calculate the moment of inertia of large ring
Using formula of moment of inertia
Where, = moment of inertia at center of mass
M = mass of ring
R = radius of ring
Put the value into the formula
Hence, The moment of inertia of large ring is 2MR².