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expeople1 [14]
3 years ago
7

If the mass of a person is 600kg on Earth what is the persons mass on the moon

Physics
1 answer:
nika2105 [10]3 years ago
3 0

Answer:

The mass of the person would remain 600kg.

Explanation:

mass refers to the amount of matter in a particular object. therefore even though the moon has less gravitational force than the Earths, it still would have the same mass since the person is made up of the same amount of matter on Earth and in the moon.

when asked about weight, it would be 1000N. but in mass , same 600kg.

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A quantity that has direction as well as size is a scalar.<br><br><br> True or False
lisov135 [29]

Answer:

A quantity that does not depend on the direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude, and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction.

Scalar quantities only have magnitude (size). Scalar quantities include distance...

A quantity that is specified by both size and direction is a vector. Displacement includes both size and direction and is an example of a vector. However, distance is a physical quantity that does not include a direction and isn't a vector.

Explanation:

hope this helps...

3 0
3 years ago
Read 2 more answers
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
Wavelengths of incoming solar radiation are __________________ the wavelengths of reradiated heat. question 2 options: faster th
MArishka [77]
I think that the wavelengths of an incoming solar radiation are shorter than the wavelengths of reradiated heat. This is because the incoming solar radiation to the surface of the earth is in the utraviolet (short) to near infrared (long) wavelength bands. After absorption has taken place, surfaces reradiate heat energy back to the atmosphere at long wavelength infrared.
6 0
3 years ago
In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a
xz_007 [3.2K]

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

u_1,u_2 = initial velocity before collision

v_1,v_2 = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

m_1=m_2=M

Initial velocity of m_1=u_1=2 m/s

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

Final velocity of m_1=v_1=v (same direction )

Final velocity of m_2=v_2=v  (same direction)

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

7 0
4 years ago
At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 25.0 rad/s^2 until
Zolol [24]

Angle turned by the wheel is given by the kinematics as

\theta = \omega t + \frac{1}{2}\alpha t^2

now here we know that

\omega = 26 rad/s

\alpha = 25 rad/s^2

t = 2.50 seconds

\theta = 26(2.50) + \frac{1}{2}(25)(2.50)^2

\theta = 143.1 rad

So here total angle that the wheel turn is given as

\theta = 143.1 + 440 = 583.1 rad

Speed of the wheel after 2.5 s

\omega = \omega_0 + \alpha t

\omega = 26 + (25)(2.5)

\omega = 88.5 rad/s

now angular deceleration is given as

\alpha_1 = \frac{\omega^2 - \omega_0^2}{2\theta}

\alpha_1 = \frac{0 - 88.5^2}{2(440)}

\alpha_1 = -8.9 rad/s^2

time taken to stop

t = \frac{\omega - \omega_0}{\alpha}

t = \frac{0 - 88.5}{-8.9} = 9.94 s

now total time is given as

T = 2.50 + 9.94 = 12.44 s

5 0
3 years ago
Read 2 more answers
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