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fiasKO [112]
2 years ago
8

We are going to focus on the gravitational force between Earth and our Moon. The Earth has a mass of about 6 x 10/24 kg and the

Moon has a
mass of about 7.3 x 10/22 kg. They are also currently about 4 x 10/8 meters apart.

The gravitational constant, G, is 6.67 x 10-/11 as always.

Question 1
If you were to weight yourself in the moon and the Earth, which would have the higher weight?

Physics
1 answer:
Basile [38]2 years ago
7 0

Answer

Gravity is what holds us down on the earth's (or moon's) surface. If you were to weigh yourself on a scale on Earth and then on the moon, the weight read on the moon would be 1/6 your earth weight

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A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0
2 years ago
A balloon is filled with helium at sea level. The volume was measured to be 3.1 liters at a pressure of 0.97 atm. If the balloon
stira [4]

Answer:

4.9L hope this helped!

8 0
3 years ago
A speed does not involve the element of
guajiro [1.7K]
A speed does not involve the element of direction.
6 0
3 years ago
I'm stuck on number 4... help please?
Paha777 [63]

Average <u>speed</u> = (distance covered) / (time to cover the distance) =
 
                                     (5m)  /  (15 sec) =

                                     (5/15) (m/s)  =  <em>1/3 m/s</em> .

Average <u>velocity</u> = 

         (displacement) / (time spent traveling)  in the direction of the displacement

Average velocity =  (5m) / (15 sec)  left =

                           (5/15) / (m/sec)  left  =

                               <em>1/3  m/s  left</em>.

A number without a direction is a speed, not a velocity.


3 0
3 years ago
Read 2 more answers
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