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Likurg_2 [28]
3 years ago
8

PLZ HELP! I WILL GIVE BRAINLIEST OR WHATEVER U PPL WANT JUST HELP!

Physics
1 answer:
Tpy6a [65]3 years ago
5 0
Newton's first law is sometimes known as the law of inertia. It is the law that states that an object at rest will stay at rest and an object in motion will stay in motion unless a force acts upon it. For example, if I was working with a wrench in space an it slipped, it would keep on going in one direction with a constant speed unless it hits something. Hope this helps!
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Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary
balandron [24]
A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
W=mg
where
m is the mass
g= \frac{GM}{r^2} is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, g=9.81 m/s^2, so Gary's weight is
W=mg=9.81 m  (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is 
R=6370 km
So the distance of Gary from the Earth's center is
r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m

The Earth's mass is M=5.97 \cdot 10^{24} kg, so the gravitational acceleration is
g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2

Therefore, Gary's weight at 300 km above Earth's surface is 
W' = mg' = 8.95 m (2)

If we compare (1) and (2), we find that Gary's weight has changed by
\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.
6 0
3 years ago
The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?
kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e           I = \frac{Q}{t} ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

            Q = I × t

                = 1.5 × 2

               = 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

5 0
3 years ago
People often use simple machines like pulleys, levers, and ramps because they say the machine “makes the work easier.” Which of
dimulka [17.4K]
I think the answer is D.
Hope this help
3 0
3 years ago
A person shouting at the top of his lungs emits aboue 1.0 w of energy as sound waves. What is the sound intensity 1.0 m from suc
umka21 [38]

Answer:

I=0.0795\ W/m^2

Explanation:

When a person shouts it emits 1 W of energy as sound waves, P = 1 W

Let I is the sound intensity 1 meters from such a person. We know that the power per unit area is called the sound intensity of the person. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{1\ W}{4\pi (1\ m)^2}  

I=0.0795\ W/m^2

So, the sound intensity of such a person is 0.0795\ W/m^2. Hence, this is the required solution.

3 0
4 years ago
Which are not in the nucleus? Check all that apply
Ivenika [448]
Ions and electrons -apex!
6 0
4 years ago
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