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3 years ago

If you weigh 637 N on Earth's surface, how much would you weigh on the planet Mars?

Physics

1 answer:

vivado [14]3 years ago

4 0

Answer:

To get your answer multiply 637N by 10N/kg to get 6370kg

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Find the mass of a person who weighs 608 N

Anvisha [2.4K]

Answer:

$weight = mass \times accelaration \: due \: to \: gravity \\ 608 = x \times 10 \\ x = 608 \div 10 = 60.8 \\ mass = 60.8 \\ thank \: you$

7 0

3 years ago

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100 Points !!!!!!!! I WILL GIVE BRAINLIEST

kumpel [21]

Answer: 1. 0.25 m/s squared

2. A = 0.4 m/s^2

3.=20kg. I did this before.

Explanation:

3 0

3 years ago

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Which statement is true of a concave lens?

natka813 [3]

It produces only virtual images is the answer

5 0

3 years ago

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A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h

Nutka1998 [239]

Answer:

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

8 0

3 years ago

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If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?

puteri [66]

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

$\omega=\sqrt{\dfrac{k}{m}}$

Put the value into the formula

$\omega=\sqrt{\dfrac{13.3}{0.321}}$

$\omega=6.43\ rad/s$

We need to calculate the velocity

Using formula of velocity

$v=\omega\sqrt{A^2-x^2}$

Put the value into the formula

$v=6.43\times\sqrt{0.270^2-0.130^2}$

$v=1.52\ m/s$

Hence, The speed is 1.52 m.

6 0

3 years ago