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alukav5142 [94]
3 years ago
13

PLEASE HELP!! I'll mark brainliest for correct answer.

Physics
1 answer:
Tju [1.3M]3 years ago
6 0
1. Car A
2. Car C
3. Car A
4. 1500 J
5. 750 J
6. 750 J
7. Car A was had a height higher relative to the ground compared to all the other cars.
8. mgh
9. B
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If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?
nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

                                                       =\frac{G4M}{16R^{2} }

                                                       =\frac{GM}{4R^{2} }

recall, gravitational force of earth is given as = \frac{GM}{R^{2} }

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0
3 years ago
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
3 years ago
Which of the following is an example of rolling friction?
il63 [147K]

Answer:

O bike tires on the road as you ride

Explanation:

is the rolling friction

7 0
3 years ago
Read 2 more answers
Identify the conditions for an elastic collision in a closed system. Check all that apply.
s344n2d4d5 [400]

Answer:

In an elastic collision:

  • There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
  • Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
  • There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
  • Velocities may change after collision. If the masses are equal, the velocities interchange.

When one object is stationary:

Final velocity of object 1:

v₁ = (m₁ - m₂)u₁/(m₁ +m₂)

Final velocity of object 2:

v₂ = (2 m₁ u₁)/(m₁+m₂) =

  • Objects do not stick together in elastic collision. They stick together in inelastic collision.
  • One object may be stationary before the elastic collision.

Thus, conditions for an elastic collision:

  • Energy is conserved.
  • Velocities may change.
  • Momentum is conserved.
  • Kinetic energy is conserved.
  • One object may be stationary before the elastic collision.
7 0
3 years ago
Read 2 more answers
Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?
bogdanovich [222]

Answer:

6 m/sec

Explanation:

12/2=6

7 0
3 years ago
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