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2 years ago

14

A

1 answer:

OleMash [197]2 years ago

7 0

Answer:

R= 5.4 ohms

Explanation:

Given that

V= 9 V

Power ,P= 15 W

Lets take resistor resistance = R

We know that Power given as

$P=\dfrac{V^2}{R}$

V=Voltage

P=Power

R=Resistance

$R=\dfrac{V^2}{P}$

Now by putting the all the values in the above equation

$R=\dfrac{9^2}{15}$

R= 5.4 ohms

Therefore the resistance of the resister will be 5.4 ohms.

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A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from

krek1111 [17]

Answer:

$E=35921.96N/C$

Explanation:

From the question we are told that:

Radius $r=0.321mm$

Charge Density $\mu=0.100$

Distance $d= 5.00 cm$

Generally the equation for electric field is mathematically given by

$E=\frac{mu}{2\pi E_0r}$

$E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}$

$E=35921.96N/C$

4 0

2 years ago

2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the colli

ss7ja [257]

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5 x v + 11 x 15

where v is velocity of A after the collision .

5 x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

3 0

2 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

2 years ago

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 2 can have? Express your answer in

butalik [34]

Answer:

$L_{max} = 1.414$ ℏ

Given:

Principle quantum number, n = 2

Solution:

To calculate the maximum angular momentum, $L_{max}$ , we have:

$L_{max} = \sqrt {l(1 + l)}$ (1)

where,

l = azimuthal quantum number or angular momentum quantum number

Also,

n = 1 + l

2 = 1 + l

l = 1

Now,

Using the value of l = 1 in eqn (1), we get:

$L_{max} = \sqrt {1(1 + 1)} = \sqrt 2$

$L_{max} = 1.414$ ℏ

4 0

2 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

2 years ago