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masya89 [10]
2 years ago
10

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

Physics
2 answers:
zalisa [80]2 years ago
6 0
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C. 
sveticcg [70]2 years ago
3 0
It would be half as strong
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A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
2 years ago
Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​
earnstyle [38]

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

6 0
2 years ago
(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
1 year ago
A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

6 0
2 years ago
In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the sprin
tankabanditka [31]

Answer:

The displacement of the spring due to weight is 0.043 m

Explanation:

Given :

Mass m = 2 Kg

Spring constant k = 450 \frac{N}{m}

According to the hooke's law,

  F = -kx

Where F = force, x = displacement

Here,

F = mg         ( g = 9.8 \frac{m}{s^{2} } )

F = 2 \times 9.8 = 19.6 N

Now for finding displacement,

  x = \frac{F}{k}

Here minus sign only represent the direction so we take magnitude of it.

  x = \frac{19.6}{450}

  x = 0.043 m

Therefore, the displacement of the spring due to weight is 0.043 m

8 0
2 years ago
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