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3 years ago

Which of two factors influence the weight of an object due to gravitational pull?

1 answer:

5 0

The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>

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How much heat is absorbed by a 28g iron skillet when its temperature rises from 7oC to 29oC?

prohojiy [21]

From the information given above,
Mass [M] = 28 g
Change in temperature = 29 - 7 = 22
Specific heat of iron = 0.449 [This value is constant]
The formula for calculating heat absorbed, Q is
Q = Mass * Specific heat of Iron * change in temperature
Q = 28 * 0.449 * 22 = 276.58 J<span />

5 0

3 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

2 years ago

Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the

Ugo [173]

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

#SPJ4

5 0

1 year ago

When the wedges are removed, the cars will move. Predict which direction they will move and when they will stop

Neko [114]

Answer: hello your question lacks the required diagram attached below is the required diagram

answer : Both cars will move backwards and stop due to friction.

Explanation:

Given that both cars are negatively charged, When the wedges are removed both cars will move backwards ( repelling each other ) because they are like poles, and Like poles repel each other. while unlike poles attract each other ( forward movement ) .

The cars will later come to a stop due to frictional forces between the cars and the surface.

6 0

2 years ago

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is

tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency $= f = 562Hz$