Tommy was walking at a rate of 4 miles hour at noon and at 12:30 pm he was walking at a brisk rate of 6 miles hour . Two hours l
2 answers:
Answer:
B) Tommy had a positive acceleration between noon and 12:30 pm.
Explanation:
Acceleration is defined as the rate of change of velocity:
where
v is the final velocity
u is the initial velocity
t is the time
In the problem,
- At noon, Tommy is walking at a velocity of 4 mi/h
- At 12.30 pm, Tommy is walking at a velocity of 6 mi/h
- A time of half an hour (0.5 h) passed between the two moments
So Tommy's acceleration is
and the acceleration is positive, since the velocity has increased.
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Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.
where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.
Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.