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3 years ago

If you take rocks from the Earth to the moon (5 points)

2 answers:

8 0

the mass would remain the same and the weight would change. mass defines the actual amount of an object, while weight is the force gravity has on that object. gravity is different on the moon and the earth therefore the weight will change

katrin2010 [14]3 years ago

4 0

Answer:

the mass would remain the same and the weight would change

Explanation:

Mass can be thought of as the amount of matter an object has, this quantity is a constant, it will not change even if the rocks are carried to the moon.

What will change is the weight, since it is defined as:

$w = mg$

where $w$ is the weight, $m$ is the mass and $g$ is the acceleration due to gravity.

We said the mass stays the same, but the acceleration $g$ on the moon will be different (smaller) than the acceleration $g$ on the earth.

So because the weight depends on the acceleration of gravity, if the rocks are taken to the moon where that acceleration is different, the rocks will have a different weight.

So the answer is: the mass would remain the same and the weight would change

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Elsie is finishing second grade. If she goes to school 147 day per year and she have 10 years of school left, how many days of s

ICE Princess25 [194]

Answer:

1,323 days left

Explanation:

147 x 10 = 1,470

1470 - 147 = 1,323

Hopefully this helps you :)

pls mark brainlest ;)

6 0

3 years ago

What is the middle layer of earth called

irinina [24]

The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle.

6 0

3 years ago

Read 2 more answers

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

grandymaker [24]

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

6 0

2 years ago

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is locat

Montano1993 [528]

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

$Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}$

Solving for Q:

$Q = \frac{m*V^{2}*r}{K*q}$

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC

5 0

3 years ago

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})$

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

5 0

3 years ago