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3 years ago

7

Two people on skateboards, Brian and Amy, push off from each other. Brian has more mass than Amy. Who experiences a larger force

?

1 answer:

amm18123 years ago

4 0

Amy because she is lighter, so the more force will be put on her

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During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

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2 years ago

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Which statement correctly describes the location, charge, and mass of the

Alenkinab [10]

The neutrons are inside the nucleus, have no charge, and have mass.

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3 years ago

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0

3 years ago

A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t

Shalnov [3]

Answer:

$C=\frac{ab}{k(b-a)}$

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

$C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\$

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

$dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]$

Hence, the capacitance is

$C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}$

but R1=a and R2=b

$C=\frac{ab}{k(b-a)}$

HOPE THIS HELPS!!

3 0

3 years ago

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To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

= 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

W = F x S joules

Substituting the values in the above equation

W = 1.5 x 10⁶ x 355 · sin θ

= 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J

6 0

3 years ago