The acceleration of gravity (on Earth) is 9.8 m/s² downward.
This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.
In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.
If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at
(2 m/s + 49 m/s) = 51 m/s .
Shiny, ductility, malleability are some
Answer:
35.14°C
Explanation:
The equation for linear thermal expansion is
, which means that a bar of length
with a thermal expansion coefficient
under a temperature variation
will experiment a length variation
.
We have then
= 0.481 foot,
= 1671 feet and
= 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a
= 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.
Since our temperature on a cool spring day was 13.0°C, our new temperature must be
= 35.14°C
Answer:
See answer
Explanation:
Given quantities:
![\eta = 0.05\\ W=90[W]\\r=0.0285[m]](https://tex.z-dn.net/?f=%5Ceta%20%3D%200.05%5C%5C%20W%3D90%5BW%5D%5C%5Cr%3D0.0285%5Bm%5D)
where
is the efficiency of the lightbulb (visible light is 5% of the total power),
is the total power of the lightbulb, r is the radius of the lightbulb in meters.
Intensity is power divided by area:

a) Now the effective power is
, therefore:
![I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]](https://tex.z-dn.net/?f=I%20%3D%5Cfrac%7B%5Ceta%2AW%7D%7B%5Cpi%20r%5E2%7D%3D%5Cfrac%7B0.05%2A90%7D%7B4%5Cpi%20%280.0285%29%5E2%7D%3D440.87%5BW%2Fm%5E2%5D)
b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:
![S_{average}= \frac{EB}{2\mu_{0}}[W/m]](https://tex.z-dn.net/?f=S_%7Baverage%7D%3D%20%5Cfrac%7BEB%7D%7B2%5Cmu_%7B0%7D%7D%5BW%2Fm%5D)
now
and
are related:
and 
replace in 
![S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]](https://tex.z-dn.net/?f=S_%7Baverage%7D%3DI%3D%20%5Cfrac%7Bc%20%5Cepsilon_%7B0%7DE%5E2%7D%7B2%7D%5BW%2Fm%5D)
we replace the values and we get:

![E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]](https://tex.z-dn.net/?f=E%20%3D%20%5Csqrt%7B%5Cfrac%7B2%28440.8%29%7D%7B8.85%2A10%5E%7B-12%7D3%2A10%5E8%7D%7D%3D576.24%5BV%2Fm%5D)
therefore
![B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]](https://tex.z-dn.net/?f=B%3D%5Cfrac%7BE%7D%7Bc%7D%3D%5Cfrac%7B576.24%7D%7B3%2A10%5E%7B8%7D%7D%3D1.92%2A10%5E%7B-6%7D%5BT%5D)
Answer:
y = 0.99 m
Explanation:
This is a projectile launching exercise, let's start by finding the components of the initial velocity, using trigonometry
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = vo cos θ
v_{oy} = I go sin θ
v₀ₓ = 15 cos 30 = 12.99 m / s
v_{oy} = 15 sin 30 = 7.5 m / s
Let's find the time it takes to travel x = 18 m
x = v₀ₓ t
t = x / v₀ₓ
t = 18 / 12.99
t = 1,385 s
at this point it is at a height of
y = v_{oy} - ½ g t²
y = 7.5 1.385 - ½ 9.8 1.385²
y = 0.99 m
therefore the camera must place the foot 99 cm from the ground