Anita is comparing the accepted value for a physical property to the value she measured in the laboratory. Which characteristic
2 answers:
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Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck
So, the acceleration of the hockey puck is .
Answer:
Final temperature will be 438.076 K
Explanation:
We have given temperature
Volume
As there is no heat transfer so this is an adiabatic process
For and adiabatic process
Here
So