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3 years ago

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Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n

itrogen under low pressure can be installed in the fuel system to check for leaks. Which technician is correct?A. Technician A only.B. Technician B only. C. Technician A and B. D. D. Neither Techncian A and B.

1 answer:

pickupchik [31]3 years ago

4 0

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

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Answer:

12

Explanation:

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3 years ago

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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

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3 years ago

The mass of the Sun is 1. 99 × 1030 kg. Jupiter is 7. 79 × 108 km away from the Sun and has a mass of 1. 90 × 1027 kg. The gravi

german

The gravitational force is s type of force that has the ability to attract any two objects having mass. The gravitational force will be $4.16\times10^{23}$ .

<h3>What is the gravitational force?</h3>

The gravitational force is s type of force that has the ability to attract any two objects with mass. Gravitational force tries to pull two masses towards each other.

$F= G\frac{m_1m_2}{r^{2} }$

Given,

mass of the sun ( $m_1$ )= $1.99\times10^{23}$ kg

mass of Jupiter( $m_2$ )= $7.79\times10^{8}$ kg

distance between the sun and Jupiter (r)= $1.90\times10^{27}$ m

$F= G\frac{m_1m_2}{r^{2} }\\\\\\F=4.16\times10^{23}\times\frac{1.99\times10^{23}\times7.79\times10^{8}}{({1.90\times10^{27})^2} }$

$F= 4.16\times(10)^{23}$ Newton

Hence the gravitational force between the sun and Jupiter will be $4.16\times10^{23}$

To learn more about gravitational force refer to the link:

brainly.com/question/24783651

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2 years ago

Who was the most famous member of the Underground Railroad?

MArishka [77]

Answer:

Harriet Tubman

Explanation:

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2 years ago

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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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2 years ago