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4 years ago

Which factors affect the strength of the electric force between two objects

2 answers:

8 0

-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)

iris [78.8K]4 years ago

5 0

By definition, the electric force is given by:

$f = k * \frac{q1q2}{d ^ 2}$

Where,

q1: electric charge of object number 1.

q2: electric charge of object number 2.

d: distance between both objects

k: proportionality constant

Therefore, the magnitude of the electric force is affected by:

1) The product of the charges of the objects

2) The distance between objects

Answer:

The factors that affect strength of the electric force between two objects are:

1) The product of the charges of the objects

2) The distance between objects

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What might happen if transformers didn’t exist?

Karolina [17]

Answer:

B electricity could not be distributed through power lines to people's homes

7 0

3 years ago

A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se

Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (3)

θ= ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

R : radius of the circular path (m)

at: tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m : radius of the disk

t₀=0 , ω₀ = 5 rev/s

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s =10π rad/s = 31.42 rad/s

ωf = 2*(2π)rad/s =4π rad/s = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0

3 years ago

Why is the sky blue in colour?

Svetllana [295]

$\huge\mathrm{Answer࿐}$

Our Atmosphere is full of freely subtended particles, and when light from sun strikes the particles then scattering occurs, in which blue colour is most scattered and hence causing Sky to appear blue .

Hope it helps !!

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$\mathrm{ \#TeeNForeveR}$

3 0

3 years ago

Read 2 more answers

A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (un

Charra [1.4K]

Answer:

1.9 s

Explanation:

We are given that

Length of cable=l=15 m

We have to find the time you have to move out of the way.

We know that

Time period,T= $2\pi\sqrt{\frac{l}{g}}$

Where g= $9.8m/s^2$

By using the formula

$T=2\pi\sqrt{\frac{15}{9.8}}$

$T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s$

Time you have to move out

$t=\frac{T}{4}=\frac{7.77}{4}=1.9 s$

Hence,time you have to move out of the way=7.77 s

6 0

3 years ago

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$