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3 years ago

Which changes will decrease the electric force between two positively charged objects? Check all that apply.

Physics

2 answers:

zaharov [31]3 years ago

8 0

Answer:

Moving them farther apart

Explanation:

The electric field between the two charges Q and q separated by a distance r is given by

$E = \frac{K\times Q\times q}{r^{2}}$

It shows that the electric field is inversely proportional to the square of the distance between two charges.

So, as the distance between two charges increases, the electric filed between the two charges decreases.

GaryK [48]3 years ago

8 0

Answer:

1. moving them farther apart

2. Adding electrons

Explanation:

The electric force between two positively charged particle is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

k is the electrostatic constant

r is the distance between charges

The electric force is inversely proportional to the distance between them. So, the electric force between two positively charged objects will decrease when the charged particles move father apart. Also, on adding electrons, the magnitude of charges will decrease.

Hence, the correct options are (b) and (c).

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Which factors could be potential sources of error in the experiment? check all that apply.

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(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

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A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .

rjkz [21]

Answer:

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Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

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3 years ago

A cars engine can deliver 300,000 watts of power to its wheels.

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Answer:

A. 1,800,000 J

B. 4473.87 N

C. 3.728 m/s²

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3 years ago

An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a

Blababa [14]

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

wave speed, $v=118\ m.s^{-1}$

<u>We have mathematical expression for strain as:</u>

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

where: $\mu=$ linear mass density of the wire

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

Using eq. (2) & (4) for tension T

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

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3 years ago

A hot-air balloon is 11.0 m above the ground and rising at a speed of 7.00 m/s. A ball is thrown horizontally from the balloon b

Nesterboy [21]

Answer:

18.6 m/s

Explanation:

$h$ = Initial height of the balloon = 11 m

$v_{o}$ = initial speed of the ball

$v_{oy}$ = initial vertical speed of the ball = 7 m/s

$v_{ox}$ = initial horizontal speed of the ball = 9 m/s

initial speed of the ball is given as

$v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{9^{2} + 7^{2}} = 11.4 m/s$

$v_{f}$ = final speed of the ball as it strikes the ground

$m$ = mass of the ball

Using conservation of energy

Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

$(0.5) m v_{f}^{2} = (0.5) m v_{o}^{2} + mgh \\(0.5) v_{f}^{2} = (0.5) v_{o}^{2} + gh\\(0.5) v_{f}^{2} = (0.5) (11.4)^{2} + (9.8)(11)\\(0.5) v_{f}^{2} = 172.78\\v_{f} = 18.6 m/s$

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