If the tow truck is pulling the car, the acceleration rate of the car is the same as the tow truck
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Answer:
D) Electromagnetic waves
Explanation:
This is because with compression waves and longitudinal waves a medium is required such as air. however in a vacuum this will not work
Answer: ![-330.47 \°M](https://tex.z-dn.net/?f=-330.47%20%5C%C2%B0M)
Explanation:
Let's begin by explaining that the relation between the Celsius scale and Kelvin scale is:
![\°C +273.15=K](https://tex.z-dn.net/?f=%5C%C2%B0C%20%2B273.15%3DK)
This means the absolute zero point of the Kelvin scale is ![0 K=-273.15 \°C](https://tex.z-dn.net/?f=0%20K%3D-273.15%20%5C%C2%B0C)
Keeping this in mind, we have the freezing point and boling point of Methane as:
Freezing point: ![0 \°M=-182.6 \°C=90.55 K](https://tex.z-dn.net/?f=0%20%5C%C2%B0M%3D-182.6%20%5C%C2%B0C%3D90.55%20K)
Boiling point: ![100 \°M=-155.2 \°C=117.95 K](https://tex.z-dn.net/?f=100%20%5C%C2%B0M%3D-155.2%20%5C%C2%B0C%3D117.95%20K)
According to this, there is a linear relation between the methane scale (
) and the Kelvin scale in the form:
(1)
Where:
is the temperature in Kelvin
is the temperature in degrees Methane
Firstly, we need to find the value of
and
with the two given points (
and
):
When
and
:
![90.55=a+b(0)](https://tex.z-dn.net/?f=90.55%3Da%2Bb%280%29)
(2)
When
and
:
![117.95=90.55+b(100)](https://tex.z-dn.net/?f=117.95%3D90.55%2Bb%28100%29)
(3)
Now we have the linear equation:
(4)
Isolating
:
(5)
Evaluating for
:
(6)
Finally:
This means
Answer:
It is (1/5)th as much.
Explanation:
If we apply the equation
F = G*m*M / r²
where
m = mass of a man
M₀ = mass of the planet Driff
M = mass of the Earth
r₀ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
F₀ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
g₀ = the gravitational acceleration on the surface of the planet Driff
we have
F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²
⇒ F₀ = G*m*M / (5*r²) = (1/5)*F
If
F₀ = (1/5)*F
then
W₀ = (1/5)*W ⇒ m*g₀ = (1/5)*m*g ⇒ g₀ = (1/5)*g
It is (1/5)th as much.
Answer:
the work done by the net external force acting on the skier is 3046.12 J.
Explanation:
Given;
initial speed of the water skier, u = 6.3 m/s
final speed of the water skier, v = 10.9 m/s
mass of the water skier, m = 77 kg
The work done by the net external force is calculated as;
W = ΔK.E
![W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28v%5E2%20-%20u%5E2%29%5C%5C%5C%5CW%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5C%2077.0%2810.9%5E2%20-%206.3%5E2%29%5C%5C%5C%5C%20W%3D%203046.12%20%5C%20J)
Therefore, the work done by the net external force acting on the skier is 3046.12 J.