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Kruka [31]
3 years ago
13

Question 4 Multiple Choice Worth 4 points)

Physics
1 answer:
guapka [62]3 years ago
7 0

Answer:

its A Weathering breaks down rocks; deposition leaves them in new places

Explanation:

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SONAR stands for "sound navigation and ranging,” and it is used to map and explore the ocean floor.
lara31 [8.8K]
SONAR stands for "sound navigation and ranging,” and it is used to map and explore the ocean floor.
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3 years ago
A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be
padilas [110]
Let
M = the mass of the planet
n = the mass of the satellite.
r = the radius of the planet

When the satellite is at a distance r from the surface of the planet, the distance between the centers of the two masses is 2r.
The gravitational force between them is
f_{0} = \frac{GMm}{(2r)^{2}} = \frac{1}{4} ( \frac{GMm}{r^{2}} )
where
G =  the gravitational constant.

When the satellite is on the surface of the planet, the distance between the two masses is r.
The gravitational force between them is
f_{4} =   \frac{GMm}{r^{2}} =4f_{0}

Answer:  f_{4} = 4f_{0}

5 0
4 years ago
Read 2 more answers
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
What happens to an object after being transported to three planets in the solar system?
alukav5142 [94]
The weight changes but the mass will stay the same.
6 0
3 years ago
An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materia
Romashka [77]

Answer:

P = 2.91*10^{-24} kg m/s

\lambda = 2.73 *10^{-10} m

size of atom hat lie in range of 1 to 5 Angstrom

\Delta x = 0.2272 Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

\lambda = \frac{h}{mv}

where h is plank constant

so\lambda = \frac{6.626*10^{-34}}{2.91*10^{-24}}

\lambda = 2.73 *10^{-10} m

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

\frac{\Delta v}{v} = 0.1

\Delta v = 0.1 v

we know that

\Delta p *\Delta x = \frac{h}{4\pi}

m \Delta v \Delta x =\frac{h}{4\pi}

\Delta x = \frac{h}{m \Delta v}

\Delta x  = \frac{2.272}{0.1}                      [\Delta v = 0.1 v]

\Delta x = 0.2272 Angstrom

7 0
4 years ago
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