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JulijaS [17]
2 years ago
9

Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.76×10^6 m, then the mass of moon will be​

Physics
1 answer:
igomit [66]2 years ago
5 0

Answer:

7.65 x 10^22kg

sorry if im wrong!

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A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0
4 years ago
A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic
tatyana61 [14]

Hello!

\large\boxed{KE = 50J}

Use the formula for kinetic energy:

KE = \frac{1}{2}mv^{2}

Plug in the given mass and velocity:

KE = \frac{1}{2} (4)5^{2}

Simplify:

KE = \frac{1}{2} (100)\\\\KE = 50 J

7 0
3 years ago
Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)
PilotLPTM [1.2K]

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

Learn more about de Broglie wavelength on

brainly.com/question/15330461

#SPJ1

6 0
2 years ago
1. What does a football player need to do to change the position of the football?​
exis [7]

Answer:

Yes, since formations aren't mentioned at all in the rules, they can be adjusted. Sometimes when making a substitution, a coach will sub in a defender for an attacker/midfielder if the team is ahead and wants to protect their lead....

Explanation:

8 0
3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
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