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Katyanochek1 [597]
3 years ago
7

Given that 1 pound is equal to 4.45 newton’s what is the weight of a 500N child in pounds?

Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

112.36 pounds

Explanation:

Since 1 pound = 4.45 Newtons, a 500N child in pounds = 500÷4.45 = 112.36 pounds (approximately).

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An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the k
laiz [17]

Answer:

10,000 eV

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the electron is equal to its change in electric potential energy:

K=\Delta U=q \Delta V

where:

K is the kinetic energy of the electron

q=1 e is the magnitude of the charge of the electron

\Delta V is the potential difference through which the electron has been accelerated

For this electron in the TV, we have

\Delta V=10 kV=10000 V

Therefore, the kinetic energy of the electron in electronvolts is

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7 0
3 years ago
a child is stationary on a swing. The child is given a push by a parent and the child starts swinging
Nesterboy [21]

Answer:

you havent given the full question

but im guessing momentum

momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity or the impetus gained by a moving object.

Explanation:

as the child is pushed, it gathers momentum as its weight allows it be pushed forward, and the velocity is the speed driven by the amount of force the parent pushes on the child whilst they are swinging. The momentum is the result of this action

the equation that links these factors together are

p = mv

p = momentum

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v = velocity

hope i got it right ._.

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2 years ago
At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
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Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

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So (86kg* 101^2)/(167) =

Fc=5253 N

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3 years ago
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