Answer:
1 - third law
2 - second law
3 - first law
4 - third law
5 - second law
6 - first law
Explanation:
First law
In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.
Second law
In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration, a of the object
F = ma.
Third law
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
Repeat the experiment
Explanation:
To increase the validity of the results obtained from the single experiment, the students should be encouraged to repeat the experiments more number of times as much as possible.
In an experiment, scientist always try to limit errors by making precise and accurate observation. A single observation does not really represent a precise and accurate finding. When an experiment is repeated as often as possible, the reliability of the conclusion drawn from the hypothesis testing will improve and the results can be accepted to be valid.
A single observation/experiment is not valid enough.
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Answer:
C. parabolic Curve
Explanation:
The trajectory of a projectile motion is parabolic.
If a ball is thrown at angle of 45° to the horizontal then it will follow a parabolic curve during its motion as its a projectile motion. The other options are incorrect as sine curve is a periodic type of curve. linear curve and tangential curve are also out of place.So option C is correct
Answer:
0.66 m/s
Explanation:
This is an inelastic collision, for that reason the conservation of momentum law allows us to determine the final velocity of both, the bullet and block together after the collision:
Answer:
Mass of ice per second melt is 2.74×10^-5Kg/s
Explanation:
Temperature of one end of the copper rod is 100°C boiling point of water and the other end of the rod is 0°C
Temperature difference in the copper rod = 100 - 0 = 100°C
Cross sectional area = 3.6×10^-4m^2
Length of rod , L = 1.7m
Amount of heat transfer from the boiling water to the ice water mix through the copper rod is given by:
Q = KA◇T/ L
Q = (390×(3.6×10^-4)×100°C)/1.7
Q = 14.04/1.7
Q = 8.26J/s
From the equation
Q = mLf
m = Q/ Lf
Where Lf = Latent heat of fusion for water= 3.34×10^5J/Kg
m = 8.26/(3.34×10^5)
m = 2.74×10^-5Kg/s
