Path length is 2*pi*0.4=2.512
Speed=distance/time
Speed =2.512/0.2=12.56m/s
Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:



Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:



Replacing the values we have:


Therefore the refractive index for the liquid is 1.7323
Answer:
which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.
This corresponds to the second option shown in the question: "Voltage times amperage".