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cluponka [151]
3 years ago
12

9. From this lab, we learn that the electric field and electric potential depend on both, the magnitude of the source charge (q)

, and the distance from the source charge (r). If we were to increase the magnitude of our source charge from 1 nC to 5 nC, then the magnitudes of the electric field and electric potential would be ____.(you can test this on the animation by dragging five 1 nC charges on top of each other and measuring E and V at a distance of 1 m)
Physics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

the electric field and the electric potential increase 5 times

Explanation:

The electric field created by a point charge is

         E = k q / r²

in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C

with the electric field is proportional to the charge

         E₅ = 5 E₁

the electrical power for a point charge is

         V = k q / r

as the electric power is proportional to the charge

         V₅ = 5 V₁

consequently both the electric field and the electric potential increase 5 times

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Seismic waves carry energy from earthquakes through the earth's crust. Seismic waves are...
Crank

Answer:

- longitudinal waves only

Explanation:

The P seismic waves travel as elastic motions at the highest speeds. They are longitudinal waves that can be transmitted by both solid and liquid materials in the Earth's interior.

3 0
3 years ago
I NEED HELP ON THIS QUESTION!
GarryVolchara [31]

Forward because things that are in motion stay in motion.

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4 0
3 years ago
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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J

8 0
4 years ago
A 215-kg load is hung on a wire of length of 3.60 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2. What
Y_Kistochka [10]

Answer:

4.74 * 10^-3

Explanation:

6 0
3 years ago
A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a
Tpy6a [65]

Answer:

The value is  F  =  34.3 \  N

Explanation:

From the question we are told that

    The mass of the shed is  m  =  25 \  kg

    The length of the rope is  l = 5.0 \  m

    The angular speed of the shed is w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \  rad/sec

Generally the force exerted on the shed is mathematically represented as

     F  =  m  *  w^2 *  l

=>  F  =  25  *  0.524^2 *  5

=>  F  =  34.3 \  N    

7 0
3 years ago
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