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Makovka662 [10]
3 years ago
14

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

e?
Physics
1 answer:
Serjik [45]3 years ago
4 0
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
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The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

8 0
2 years ago
Jaime lifts a package weighing 75N. if she lifts it 1.2 m, what work has she done
9966 [12]
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
6 0
3 years ago
A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only
ANTONII [103]

Answer:

Technician  A  is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R  the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split  between the two bulbs ( we  could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0
3 years ago
What causes one plate to sink under another <br> 1 Mass<br> 2 Shape<br> 3 Density
solong [7]
It is either the mass or density, I believe it is 3 density tho but it could be mass good luck:)
3 0
2 years ago
T or F :)
balandron [24]

That's true.

And I'll go ya a better one:

If the object is moving or not moving, at a constant or changing speed, in a straight or curvy line, and the forces on it do not cancel out and add up to zero, the object will accelerate.

4 0
3 years ago
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