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3 years ago

8

A lightning bolt has a current of 56,000 A and lasts for 80 x 10-6 seconds (80 μs). How much charge (in Coulombs) has flowed in

this bolt?

1 answer:

Alina [70]3 years ago

8 0

Answer:

A cloud can discharge as much as 20 coulombs in a lightning bolt.

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

3 years ago

(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this soleno

Paul [167]

Answer:

a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H

Explanation:

a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.

Also, Li = Ф where L = inductance of solenoid.

So, Li = NAB

= NA(μ₀ni)

= NA(μ₀Ni/l)

Li = μ₀AN²i/l

dividing both sides by i, we have

So, L = μ₀AN²/l

b. The self- inductance, L = μ₀AN²/l where

A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m

So, L = μ₀AN²/l

L = μ₀πd²N²/4l

L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)

L = 11,103.3 × 10⁻¹¹ H

L = 1.11033 × 10⁻⁷ H

L ≅ 1.11 × 10⁻⁷ H

6 0

3 years ago

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

2 years ago

A boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radiu

irina [24]

Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

ω² = 926

ω = 30.4 rad /s

angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

time taken = θ / ω

= 126.6 / 30.4

= 4.16 s .

3 0

3 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

2 years ago