That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.
<span>(a) How fast is it moving when it reaches 12.0 m?
To determine the velocity as it reaches 12.0 m, we use one of the kinematic equations,
</span>V^2 = Vo^2 + 2gh
<span>where Vo = 20 m/s. </span>
<span> g = -9.8 m/s^2 </span>
<span> h = 12.0 m. </span>
V^2 = 20^2 + 2(-9.8)(12.0)
<span>V^2 = 164.8
V = 12.84 m/s
(b) How long is required to reach this height?
To determine the maximum height, we use the same equation we used above,
</span>V^2 = Vo^2 + 2gh
where Vo = 20 m/s.
g = -9.8 m/s^2
V = 0 (since at the maximum height velocity is zero)
0^2 = 20^2 + 2(-9.8)h
<span>h = 20.41 m
(c) Why are there two answers for (b)?
There are two answers for b because it would travel a distance up and travel a distance down.</span>
1 mole of photons contain

photons (Avogadro number). This means that 3.0 moles of photons contain

photons.
The wavelength of the light in the problem is

, so the frequency is

The energy carried by a single photon is

where

is the Planck constant, while f is the frequency. Since this is the energy carried by a single photon, the energy carried by 3.0 moles of photons will be the energy of the single photon multiplied by the total number of photons:

which corresponds to E=536 kJ.
Answer:
Distance, d = 3242.19 meters
Explanation:
It is given that,
Speed of the dog, v = 19.85 mi/hr
Since, 1 mph = 0.44704 m/s
v = 19.85 mi/hr = 8.873 m/s
Time taken by the dog, t = 6.09 min = 365.4 sec
Let the distance covered by the dog during this time period is d. It can be calculated by the speed of the dog multiplied by the time taken as :


d = 3242.19 meters
So, the the dog travel during this time period is 3242.19 meters. Hence, this is the required solution.
Answer:
dosimetry
Explanation:
dosimetry is the determination of energy imparted to matter by radiation