Answer:
Av = 25 [m/s]
Explanation:
To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.
where:
Av = speed [km/h] or [m/s]
distance = 180 [km]
time = 2 [hr]
Therefore the speed is equal to:
![Av = \frac{180}{2} \\Av = 90 [km/h]](https://tex.z-dn.net/?f=Av%20%3D%20%5Cfrac%7B180%7D%7B2%7D%20%5C%5CAv%20%3D%2090%20%5Bkm%2Fh%5D)
Now we must convert from kilometers per hour to meters per second
![90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]](https://tex.z-dn.net/?f=90%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A1000%5B%5Cfrac%7Bm%7D%7B1km%7D%5D%2A1%5B%5Cfrac%7Bh%7D%7B3600s%7D%20%5D%3D%2025%20%5Bm%2Fs%5D)
Answer:
a) α=7.9x10^-4 rad
b) θ=1.12x10^-4 rad
c) The Earth and the Moon cannot be seen without a telescope.
Explanation:
In this exercise we will use the concepts of angular resolution, which depends on both the wavelength of the rays and the diameter of the eye or lens on the meter. Its unit of measure is the radian. The attached image shows the solution step by step.
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
<u>τ = 4900 N.m</u>
<u></u>
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
<u>τ = 4900 N.m</u>
<u></u>
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
<u>τ = 4900 N.m</u>
<u></u>
Hence, the correct answer will be:
<u>e. The torque is the same for all cases.</u>
the greater the <u>mass</u> of an object the more force is needed to cause acceleration
