You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too
l is pushed on a frictionless horizontal surface by a 12.7 n force, it moves 16.1 m in the first 2.50 s , starting from rest. you next observe that if you release this tool from rest at 10.3 m above the ground, it takes 2.88 s to reach the ground.
1 answer:
Refer to the diagram shown.
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
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Answer:
1.
2.
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.
Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time
Now that acceleration is halved:
You would need to push for twice amount of time
2. The distance traveled by the puck is as the following equation:
So if the acceleration is halved while maintaining the same d:
As , then
. Also
So t increased by 1.14
Answer:
option (c) is correct
Explanation:
Half life of a substance is the time in which the element becomes half of is initial value.
half life, T = 8 days
Amount remaining, N = 10 % of original value
Let the original value is No.
N = 10% of No
N = 0.1 No
Let the time taken is t and the decay constant is λ.
The relation between the decay constant and the half life is given by
Us the equation of radioactivity
Taking natural log on both the sides, we get
0.08664 t = 2.303
t = 26.6 days