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3 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

Physics

1 answer:

il63 [147K]3 years ago

4 0

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

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An astronaut in her space suit has a total mass of 87.0kg including suit and oxygen tank. Her tether line loses its attachment t

Shtirlitz [24]

Answer:

Explanation:

a )

In space due to weightlessness both astronaut and her oxygen tank will float .

when she throws the tank away from spacecraft , she will have a velocity in opposite direction ie towards the spacecraft . This happens due to conservation of momentum . She creates a momentum away so that she can get a momentum towards the spaceship.

m₁ v₁ = m₂v₂

12 x 8 = ( 87 - 12 ) x v₂

v₂ = 1.28 m /s

Time allowed = 2 x 60

= 120 s

So maximum distance upto which she can remain away from spacecraft

= 120 x 1.28

= 153 m .

b )

The Newton's law which explains the theory behind it is "third law of motion" . This law gives law of conservation of momentum .

6 0

4 years ago

A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.7 m/s2. (a) if the tension in the cord is 63

zubka84 [21]

(a)
The formula is:
∑ F = Weight + T = mass * acceleration

as the elevator and lamp are moving downward, I choose downward forces to be positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2

(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides

(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63

mass = 63 / 11.5

Mass = 5.48 kg

(b)
Since the elevator and lamp are going upward, I choose upward forces to be positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7

-53.70 + T = 5.48 * 1.7

T = 53.70 + 9.316 = approx 63 N

The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated upward at the same rate of 1.7 m/s^2

3 0

3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

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

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R 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 velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

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4 years ago

When plate movement rocks to break its called

Lilit [14]

It is called a fault

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3 years ago

What is the acceleration given to a 50-kg block of cement when it is pulled

melomori [17]

the answer is a= 16m/s2

you can use the formula f=m*a to solve this bb :)

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4 years ago

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