PHYSICS MIDDLE SCHOOL

An object's weight is dependent upon its location in the universe. Why is this true?A. This is true because gravitational force is the magnitude of weight acting on an object's mass and since weight varies in different places in the universe, both gravitational force and mass also vary.

B.This is true because weight is the amount of matter in an object. Because gravity is a force, and force is proportional to weight, both gravity and weight vary in different places in the universe.

C. This is true because gravitational force is the magnitude of weight acting on an object and since weight varies in different places in the universe, gravitational force also varies.

D. This is true because weight is the magnitude of gravitational force acting on an object, and since gravitational force varies in different places in the universe, weight also varies.

Answers

Answer 1

Answer:

Explanation :

We know that the weight of an object is given by :

$W=mg$

Where,

m is the mass

g is the acceleration due to gravity.

The value of g changes from one location to another.

An object's weight is dependent upon its location in the universe. This statement is true. This is because g varies in different places so the weight also varies.

Hence, the correct option is (D) " This is true because weight is the magnitude of gravitational force acting on an object, and since gravitational force varies in different places in the universe, weight also varies ".

Answer 2

Answer:
Because an object's weight is the measure of the gravitational
forces between the object and other things that have mass.

In its travels around the universe, the object can be in the
neighborhood of other things with huge mass or tiny mass,
and it can be nearer to them or farther away from them. So
the gravitational forces between the object and other things
can have widely different values in different places.

Read the choices very carefully.
I think 'D' is the one that says this.

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Do I have to be good at math to be a physicist?

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Yes, It does require some math skills.

Give one example of a protista.

Answers

Plasmodium falciparum

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Barometers have long been used to predict the weather. Jay and Jeff were responsible for recording the class weather data each day in March. One day, they noticed that there was a rapid change in the barometric pressure. The barometric pressure dropped down very, very quickly. What type of weather could Jay and Jeff expect soon? A) fog B) rain C) clearing D) drop in temperature

Answers

Answer:

B) rain

Explanation:

The barometer is a device that is used to measure atmospheric pressure and, consequently, the height at which someone rises, as well as to approximately predict atmospheric changes. Thus, a rising barometer indicates an increase in air pressure, whereas a falling barometer indicates a decrease in air pressure.

In environments with low atmospheric pressure and, therefore, heat, the heated air tends to rise and, when reaching higher altitudes, condenses, forms clouds and precipitates. In this situation, the barometric pressure drops very, very quickly, indicating that a rain is coming.

Answer:

Explanation:

You would have rain. You would have clouds, precipitation, and strong winds.

Earth has a mass of 5.97 x 1024kg, and a mean radius of 6.38 x 10m. What would be the orbitasatellite in orbit 1.44 x 108 m above Earth?
a. v, = 630m/s,
c. v, - 1630m/s,
T = 2.78 x 10 s
T=5.78 x 10s
b. V = 1820m/s,
d. V, - 1260m/s,
T = 6.78 x 106 s
T=5.78 x 10's
Please select the best answer from the choices provid

Answers

Answer:

v = 1630 m/s

T = 5.78 x 10^5 s

Explanation:

The tangential speed of the satellite can be found by requiring that the gravitational force on the satellite is equal to the centripetal force:

$G(Mm)/((R+h)^2)=m(v^2)/(R+h)$

where

G is the gravitational constant

M=5.97 x 1024kg is the Earth's mass

m is the satellite's mass

$R=6.38 \cdot 10^6 m$ is the Earth's radius

$h=1.44\cdot 10^8 m$ is the altitude of the satellite

v is the speed of the satellite

Solving for v,

$v=\sqrt{(GM)/(R+h)}=\sqrt{((6.67\cdot 10^(-11))(5.97\cdot 10^(24)kg))/(6.38\cdot 10^6 m+1.44\cdot 10^8 m)}=1627 m/s \sim 1630 m/s$

And the period of the orbit is equal to the ratio between the distance covered during one revolution (the circumference of the orbit) and the speed:

$T=(2 \pi (R+h))/(v)=(2\pi (6.38\cdot 10^6 m+1.44\cdot 10^8 m))/(1630 m/s)=5.79\cdot 10^5 s$

So the correct answer is

v = 1630 m/s

T = 5.78 x 10^5 s

Newton discovered that gravity behaves differently in space than it does on Earth. true or false?

Answers

Yes. It is true. Gravity behaves differently than on earth. In-fact, as we go above the earth surface the gravity decreases. Gravity as we go above the earth surface can be calculated using following formula:

g' = g(1-2h/R)

Hence, it is true that the gravity behaves differently than on earth.

an object at rest starts accelerating if it travels 30 meters to end up going 10 m/s what was it’s acceleration