Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

So now we have an equation and unkown value.
for the thrown rock

for the dropped rock

solving both equation with the quadratic formula:

we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
Answer:
hi here is your answer and this is a very important question.
Explanation:
A lever is a rigid bar with three parts: the fixed point around which the bar pivots is the fulcrum: the effort arm (in-lever arm) is the part of the lever to which force is applied; the resistance arm (out-lever arm) is the part that bears the load to be moved.
Answer:
F = f from Newton’s first law.
Explanation:
since the desk is moved in a straight line at a constant speed, newton first law tell us that the two forces must be equal.
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. since the table has been set in motion by the 400 N force, it will remain in motion unless it is been acted upon by an external force, and this means that the 400 N must be equal to the frictional force for it to have been in motion in the first instance.
Answer:
W = 3.12 J
Explanation:
Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:
β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C
So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

So ΔV = 3.0843*10^-5 m^3
Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa
To get work, multiply the air pressure and the volume change.

W = 3.12 J
Hope this helps!
Answer:
527 Hz
Solution:
As per the question:
Beat frequency of the player, 
Frequency of the tuning fork, f = 523 Hz
Now,
The initial frequency can be calculated as:


when

when

But we know that as the length of the flute increases the frequency decreases
Hence, the initial frequency must be 527 Hz