Which mass is undergoing to the greatest amount of acceleration ??
1 answer:
Answer:
Option (3)
Explanation:
Formula used to calculate acceleration is,
F = ma
Where F = force exerted on a mass
m = mass
a = acceleration due to force exerted on the mass
Option (1),
When F = 100 N and m = 100 kg
100 = 100a
a = 1 m per sec²
Option (2)
For F = 1 N and m = 100 kg
1 = 100a
a =
a = 0.01 m per sec²
Option (3)
For F = 100 N and m = 1 kg
100 = 1(a)
a = 100 m per sec²
Option (4)
For F = 1 N and m = 1 kg
1 = 1(a)
a = 1 m per sec²
Therefore. acceleration in Option (3) is the maximum.
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<h3>Question:5</h3>
The given data can be written in following way in coordinates axes.
(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).
a) Average velocity for first 4 seconds
Average velocity = Total Displacement/ Time taken
= (4-0)/(4 - 0)
=4/4
= 1/1
<h3> =1 m/s</h3>b) Average velocity for 4 to 8 seconds
Average velocity = Total Displacement/ Time taken
= (4 - 4)/(8-4)
= 0/4
<h3> = 0</h3>c) Average velocity for last 6 seconds
Last 6 seconds = from 8 to 14 seconds
Average velocity = Total Displacement/Time taken
= (2 - 4)/(14 - 8)
= -2/6
<h3> = -1/3 m/s</h3><h3>Question 6:</h3>The given data can be written in following way in coordinates axes.
(0,0), (10,20), (20,20), (30,20), (40,0)
a) State the kind of motion from Os to 10s and from 30s to 40s
It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.
b) What is the velocity of the body after 10s and 40s ?
It is clear from the graph and table as well that,
<h3>Velocity after 10s is 20m/s</h3>and
<h3>Velocity after 40s is 0 m/s</h3>c) Calculate the distance covered by the body between 10s to 30s.
Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD
Area of rectangle ABCD= (30 - 10) × (20 - O)
=20×20
<h3> = 400 m</h3>Answer:
electric potential is 3.31 × V
potential energy is 152 MeV
Explanation:
given data
fragment charge Q = 46 protons = 46 × 1.6 × C
to find out
electric potential and potential energy
solution
we know here distance from fragment d = 2 × m
and constant for electric force k that is 9 × N-m²/C²
so that we can find electric potential = kQ/d
electric potential = 9 × / ( 2 ×
)
electric potential = 3.31 × V
and
we know relation between electric potential and potential
that is V = U/q
so U will be = qV
now put all value
we get potential energy U
potential energy = 46 × 3.31 ×
potential energy = 1.52 × eV
so potential energy = 152 MeV
Answer:
The magnitude of the average angular acceleration is calculated as
Explanation:
Maximum speed that can be attained by the disk, = 10,000 rpm
Speed of spinning of the disk, N = 7570 rpm
Time taken to come to rest, t = 0.435 s
Now,
The initial angular velocity is given by:
Final angular velocity,
The average angular acceleration of the disk can be computed by using the kinematic eqn: