Which one of the following does not accurately describe the universal gravitational law? A. The object with the smaller mass doe
2 answers:
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1. • Here, force of gravity on the block = 20 N.
• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]
• The coefficient
• Force of friction =
• Hence, the force of sliding friction between the block and the ground is 8 N.
• So, it is option c. 8 N
2. The answer is option d. continue in the same direction with no change in speed.
We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.
3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.
Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.
So, 1 N = 1 Kg × 1 m/s^2.
4. It is d. not zero.
Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.
5. Force = 2 N
Acceleration of the object A = 2 m/s^2.
Acceleration of the object B = 1 m/s^2.
Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg
And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg
So, the mass of object B is greater than that of object A.
Hence, the answer is option c. Object B has more mass.
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
speed of golf ball is 1.15 × m/s
and % of uncertainty in speed = 2.07 × %
Explanation:
given data
mass = 45.9 gram = 0.0459 kg
speed = 200 km/hr = 55.5 m/s
uncertainty position Δx = 1 mm = m
to find out
speed of the golf ball and % of speed of the golf ball
solution
we will apply here heisenberg uncertainty principle that is
uncertainty position ×uncertainty momentum ≥ ......1
Δx × ΔPx ≥
here uncertainty momentum ΔPx = mΔVx
and uncertainty velocity = ΔVx
and h = 6.626 × Js
so put here all these value in equation 1
× 0.0459 × ΔVx =
ΔVx = 1.15 × m/s
and
so % of uncertainty in speed = ΔV / m
% of uncertainty in speed = 1.15 × / 55.5
% of uncertainty in speed = 2.07 × %