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Dominik [7]
3 years ago
15

Calculate the gravity field at the surface of the Sun.

Physics
1 answer:
Alenkinab [10]3 years ago
7 0

Answer:

The force of gravity is the gravitational constant multiplied by the mass of the sun times the mass of the person divided by the radius of the sun squared. We can equate these two. The mass of the person cancels.

You might be interested in
Three 20.0 ohm resistors are
V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0
3 years ago
Based on its orbit, which planet behaves the least like the others?
Gekata [30.6K]
I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps
3 0
3 years ago
At what condition does a body becomes weightless at the equator?
ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0
2 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
A certain copper wire has a resistance of 10.5 ω. at what fraction of the length l must the wire be cut so that the resistance o
Delvig [45]
The answer for that questions will be 759. Stars itir
6 0
3 years ago
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