Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
Answer:
I believe the answer is B.
Explanation:
Newton's First Law of Gravity states, "The greater the weight (or mass) of an object, the more inertia it has. Heavy objects are harder to move than light ones because they have more inertia.
"
Option B The thickness of the central portion of a thin conveying lens can be determined very accurately by using a micrometer screw gauge.
<h3>What can be measured using a micrometer screw gauge?</h3>
One micrometer of thickness can be measured with a micron micrometre screw gauge. A Use of Micrometer Screw Gauge as like example Upon turning the screw of the micrometer screw gauge four times, a 2 mm space is covered.
<h3>What purposes does a micrometer serve?</h3>
A tool known as a micrometer is used to measure solid objects’ lengths, thicknesses, and other dimensions precisely and linearly.
<h3>What is the micrometer screw gauge’s SI unit?</h3>
The SI symbol m is also known as a micron, which is an SI-derived unit of length equaling 1106 meters, where 106 is the SI standard prefix for the prefix “micro-.” A micrometer is one-millionth of a meter.
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The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.
Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg
F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)
F = 83 N
Answer:
A. 
B. 
C. 
Explanation:
Given:
- spring constant,
- mass attached,
A)
for a spring-mass system the frequency is given as:
B)
frequency is given as:
C)
Time period of a simple harmonic motion is given as:
