A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring
1 answer:
Answer:
4. both blocks will both have the same amount of kinetic energy.
Explanation:
When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.
Spring force is given as:
where: length of compression in the spring
<u>We know kinetic energy is given by:</u>
Hence the kinetic energy of both the blocks is equal when they are released to move free.
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Answer:
461 km/h
Explanation:
In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).
We have two situations there, the first one is Mustapha's car that is traveling at a constant speed of 231km/hr.
The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/.
We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.
So let's model the first situation.
The displacement of Mustapha can be found by using the following equation:
when solving the equation for the displacement x we get that it will be:
Now let's model the displacement of the cop. Since the cop has a constant acceleration, we can model his displacement with the following formula.
Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:
We can now set both equations equal to each other so we get:
When solving this for t, we get that:
(let's call this equation 1)
Now, we know the cop has constant acceleration, so we can model it with the following formula too:
since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:
when solving for the final velocity, we get that:
(let's call this equation 2)
when substituting equation 1 into equation 2 we get:
we can now cancel a leaving us with:
This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:
so
which is our answer.
Answer:
All the answers are solved and explained below.
Explanation:
Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.
Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).
Data Given:
m = 1000 kg
F = 2000N
t = 10s
Q1 Solution:
Acceleration = a = ?
F = ma
a = F/m
a = 2000/ 1000
a = 2
Q2: Solution:
Change in velocity = Δv = ?
acceleration = change in velocity / time
a = Δv/t
Δv = axt
Δv = 2 x 10
Δv = 20 m/s
Q3: Solution:
Impulse = I = ?
Impulse = Force x time
I = 2000 x 10
I = 20000 Ns
Q4: Solution:
Change in Momentum = Δp = ?
Δp = mΔv
Δp = 1000 x 20
Δp = 20000 Kgm/s
Q5: Solution:
Final velocity of the car at the end of 10 seconds = vf = ?
Δp = m x Δv
Δp = m x (vf-vi)
Δp = 1000 x (vf - 0 )
20000 = 1000 x vf
vf = 20000/1000
vf = 20 m/s
Q6: Solution:
Change in momentum the car experiences as it continues at this velocity?
Δp = ?
Δp = mΔv
Δp = m x (0)
Δp = 0
Q7: Solution:
Impulse = Change in momentum
Impulse = Δp
Implulse = 0
Q8: Solution:
Change in momentum = Δp = mΔv
Δp = m(vf-vi)
Δp = 1000 x (0-20)
Δp = -20000 kgm/s
Q9: Solution:
Impulse = Δp
Impulse = -20000 Ns
Q10: Solution:
Impulse = ?
Impulse = F x t
F = impulse/t
F = -20000/4s
F = -5000 N
Q11: Solution:
F = ma
a = ?
a = F/m
a = -5000/1000
a = -5
Answer:
The change in height of the mercury is approximately 2.981 cm
Explanation:
Recall that the formula for thermal expansion in volume is:
from which we solved for the change in volume due to a given change in temperature
We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:
Therefore, the change in volume with a change in temperature of 36°C becomes:
Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):