5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o
1 answer:
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We are given: Final velocity ()=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion
Let us plug values of final velocity, and time in above equation.
Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion
Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,
Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
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:
Electric field strength= Force/unit charge
E= (kQq/r²)/q ₓ r
where r is the unit vector in the direction of unit charge
E=