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State any two effects of gravitational force

Connection to Big Idea about energy: Gravity creates gravitational potential energy. Gravitational energy relies on the masses of two bodies and their distance.

Connection to Big Idea about the universe: Gravitational force is exerted by all objects with mass throughout the Universe. It is what keeps the Earth and the planets in orbit around the Sun, and our Solar System in orbit around the centre of the Milky Way. Gravity is one of the forces involved in the birth of stars, their evolution and finally their death.

Connection to Big Idea about Earth: The gravitational force is responsible for many physical properties of Earth and consequently it affects the existence and the properties of living creatures on it. For instance, the existence, the chemical composition and the structure of Earth’s atmosphere was determined by Earth’s gravitational force.

3 0

3 years ago

On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea

snow_tiger [21]

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

4 0

3 years ago

The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60 cm ) sin ( 6.20 y+ 60.0 t ), where y is in m

trapecia [35]

Answer:

The question is incomplete, below is the complete question

"The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ), where y is in m and t is in s.

A) What is the frequency of this wave?

B) What is the wavelength of this wave?

C) What is the speed of this wave?"

Answers:

a. $f=\frac{30}{\pi }Hz\\$

b. $wavelength=\frac{\pi }{3.1}m \\$

c. $v=9.68m/s$

Explanation:

The equation of a wave is represented as

$D(x,t)=Asin(kx+wt) \\$

Where A=amplitude

w=angular frequency=2πf

K=wave numbers =2π/λ

since we re giving he equation D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ),

we can compare and get the value for the wave number and angular frequency.

By comparing we have

w=60rads/s

k=6.20

a. to determine the frequency, from the expression fr angular wave frequency we have

w=2πf hence

f=w/2π

if we substitute we arrive at

$f=\frac{60}{2\pi }\\f=\frac{30}{\pi }Hz\\$

b. to determine the wave length, we use

$k=\frac{2\pi }{wavelength} \\k=6.2\\wavelength=\frac{2\pi }{k} \\wavelength=\frac{2\pi }{6.2} \\wavelength=\frac{\pi }{3.1}m \\$

c. the wave speed v is express as the product of the frequency and the wavelength. Hence

$v=frequency*wavelength \\v=\frac{30}{\pi } *\frac{\pi }{3.1}\\ v=9.68m/s$

6 0

3 years ago

A rock is dropped from a vertical cliff. The rock takes 6.00 s to reach the ground below the cliff. What is the height of the cl

Lana71 [14]

Answer:

180 m

Explanation:

The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation

$h=\frac{1}{2}gt^2$

where

g = 10 m/s^2 is the acceleration due to gravity

t = 6.00 s is the time of the fall

Substituting these data, we find the height of the cliff:

$h=\frac{1}{2}(10 m/s^2)(6.00 s)^2=180 m$

4 0

4 years ago

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

xeze [42]

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ

6 0

4 years ago