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3 years ago

How does an inclined plane trade force for distance??

1 answer:

8 0

The sloping surface of the inclined plane supports part of the weight of the object as it moves up the slope. As a result, it takes less force to move the object uphill. The trade-off is that the object must be moved over a greater distance than if it were moved straight up to the higher elevation.

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Garret's swimming coach is giving him important and helpful tips to improve his swimming skills. Which gesture suggests that Gar

bogdanovich [222]

Answer: "important and helpful" Show garret is paying close attention to her coach. Hope this help!

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3 years ago

Read 2 more answers

Why is a warm, tropical cumulus cloud more likely to produce precipitation than a cold, stratus cloud?

Lelu [443]

Explanation:

Warm,tropical cumulus cloud more likely to produce precipitation because they have high liquid content, strong and consistent updraft, these clouds are very thick. Moreover, they have large range of cloud droplet sizes. Whereas, stratus cloud have somewhat opposite characteristics, hence tropical cumulus clouds produce more precipitate.

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3 years ago

Four students are each pushing on one side of a box. Mara is pushing to the west, Cindi is pushing to the east, Chris is pushing

jenyasd209 [6]

Answer:

cindi

Explanation:

cindi's work done is larger than all the other students combined

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3 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

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3 years ago

An electric charge, A, is placed carefully between two other charges, B and C, and experiences no net electric force. Do B

Brrunno [24]

Answer:

I do not have enough information to tell

Explanation:

This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.

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3 years ago