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creativ13 [48]
3 years ago
13

PLEASE SOMEONE HELP ME ILL GIVE THE PERSON BRAINLIEST

Physics
1 answer:
Yuliya22 [10]3 years ago
5 0
I believe it would be c
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By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)
Fofino [41]

Gravity on the surface of sun is given as

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

here we know that

M = 1.98 \times 10^{30} kg

R = 6.95 \times 10^8 m

now we will have

g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}

g = 273.4 m/s^2

now we need to find the ratio of weight on surface of sun and on surface of Earth

R = \frac{mg_{sun}}{mg_{earth}}

R = \frac{273.4}{9.8} = 28 times

so weight will increase by 28 times

6 0
2 years ago
1) How is density different than mass?
elixir [45]
Mass is a measure of how much matter there is within an object, typically given in grams. Mass is not affected by gravity, so a given object would have the same mass on earth as in outer space. Density is the amount of mass in an object per a certain amount of volume
3 0
3 years ago
A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric
vichka [17]

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

4 0
3 years ago
Read 2 more answers
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len
Mekhanik [1.2K]

Answer:

  • The difference in length for steel is 2.46 x 10⁻⁴ m
  • The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, \gamma_{st.} =  3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, \gamma_{in.} =  2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂  = L₁ + L₁αθ

L₂  - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = \frac{\gamma}{3}

Difference in length, for steel at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC

ΔL  = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0
3 years ago
Move the red arrow onto the picture to show how you think warm air would move over this area of land and water on a dark cold ni
Bess [88]
There’s no pictures?
7 0
2 years ago
Read 2 more answers
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