Answer:
Canyon is 50.176 meter deep.
Explanation:
The students is standing on the rim of the canyon and drops down a rock from the rim(cliff). We have to find what is the depth of the canyon i.e. how much below is the ground from the cliff.
Given data:
Time = t = 3.2 s
Initial velocity =
= 0 m/s
Gravitational acceleration = g = 9.8 m/s²
Height = h = ?
According to second equation of motion

As initial velocity is zero, So the first term of right hand side of above equation equal to zero

h = (0.5)(9.8)(3.2)²
h = 50.176 m
This means, the rock traveled 50.176 meters to reach the bottom of the Canyon. So, the Canyon is 50.176 meter deep.
The spoon to transfer 40 J of energy to your hand is descibed as follows
<u>Explanation:</u>
Given area of cross section of copper spoon is A = 20mm into 1.5 mm
temperature difference is DT = (100 minus 35) = 65 0C
length of the spoon is l = 18 cm,
amount of heat should be transfer Q = 40 J
coefficient of thermal conductivity of copper k = 400 W by mk
we know that the thermal conductivity is Q by t = k into A into DT by l
t = Q into l by k into A into DT
t = (40 into 0.18) by 
t = 9.23 s
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.
First of all, we need to calculate the speed of sound at temperature of

:

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):

where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
Answer:

Explanation:
<u>Free Fall Motion</u>
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:

Where 
Similarly, the distance y the object has traveled is calculated as follows:

If we know the height h from which the object was dropped, we can solve the above equation for t:

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

Therefore, it has traveled down a distance:

Thus, the height of the pen is:

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.