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steposvetlana [31]
2 years ago
10

The electric field strength at a point in space a fixed (known) distance from a point charge source is 5000 N/C. How does the fi

eld strength change if the source charge has doubled charge
Physics
1 answer:
Romashka-Z-Leto [24]2 years ago
7 0

Answer:

See explanation

Explanation:

Electric field strength;

E = kq/d^2

k= coulombs constant

q= magnitude of charge

d = distance if separation

If

d is constant

q = 2q

Then;

E = 2 kq/d^2

Hence, the electric field strength will double.

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The answer is 4000N...
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3 years ago
a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resi
777dan777 [17]

Answer:

<h2>d₂ = 3d</h2><h2>The diameter of the second wire is 3 times that of the initial wire.</h2>

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

5 0
3 years ago
An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and
Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

\Delta P.E=mg(h_{2}-h_{1})

\Delta P.E=10000\times9.8\times(10000-0)

\Delta P.E=10000\times9.8\times10000

\Delta P.E=980000000\ J

\Delta P.E=980\ MJ

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)

\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)

\Delta K.E=148298642\ J

\Delta K.E=148.3\ MJ

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0
3 years ago
An aircraft has a mass of 3.30 x 105 kg. At a certain instant during its landing, its speed is 45.2 m/s. If the braking force is
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How much heat is required to warm 122 g of water by 23.0 c?
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<span>122 g * 4,186 (j/g*°c) * 23°c = 11745.916 j </span>
4 0
3 years ago
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