The symbols for elements have either

5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

lakkis [162]

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

3 0

3 years ago

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago

The power in an electric circuit varies inversely with the resistance. If the power is 2,200watts when the resistance is 25 ohms

lana66690 [7]

Answer:

50ohms

Explanation:

If the power in an electric circuit varies inversely with the resistance, it can expressed as P = k/R where;

P is the power, R is the resistance and k is the constant of proportionality.

From the formula, k = PR

k = P1R1 = P2R2...(1)

Given P1 = 2200watts, R1 = 25ohms, P2 = 1100watts R2 = ?

From equation (1);

R2 = P1R1/P2

R2 = 2200×25/1100

R2 = 50ohms

The resistance when the power is 1100watts is 50ohms.

This shows that as the power is decreasing, the resistance is increasing.

8 0

3 years ago

Read 2 more answers

When velocity is graphed with respect to time, what does the area under

Juliette [100K]

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

4 0

3 years ago

Read 2 more answers

A potential energy function is given by U(x)=(3.00J)x+(1.00J/m^2)x^3. What is the force function F(x) (in newtons) that is assoc

snow_lady [41]

Answer:

$F(x)=-3 N -(3N/m^2) x^2$