Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.
v is the speed of cat,
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
D. Molecules of a gas slow down and change to a liquid state.
Explanation:
- Condensation refers to a process by which a gas changes from gaseous state to liquid state. For example, water vapor changes to from the state of being a gas to liquid state water.
- Condensation is the opposite of evaporation and occurs when gaseous particles slow down and change into liquid state.
- Heat energy is lost during condensation and gaseous molecules lose kinetic energy making them to slow down and thus changing to liquid state,
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Answer \|/
Ice is less dense than water.
Reason why \|/
When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats
Explanation:
Let is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :
r is the radius of path,
Time period is given by :
Frequency of proton is given by :
The wavelength of radiation is given by :
So, the wavelength of radiation produced by a proton is . Hence, this is the required solution.
Answer:
8.8 m and 52.5 m
Explanation:
The vertical component and horizontal component of water velocity leaving the hose are
Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level
t = 3.49 or t = 0.58
We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down
t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building