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3 years ago

10

A solenoid with 435 turns has a length of 7.50 cm and a cross-sectional area of 3.50 ✕ 10−9 m2. Find the solenoid's inductance a

nd the average emf around the solenoid if the current changes from +4.00 A to −4.00 A in 6.83 ✕ 10−3 s.

1 answer:

OLga [1]3 years ago

4 0

Answer:

Solenoid's inductance is 1.11 × 10^-8H

The average emf around the solenoid is 1.3 × 10^-5V

Explanation: Please see the attachments below

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A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50m/s . The pulling force is 100 N parallel to t

sergey [27]

167.67 Joules of work are done by the gravitational force on the crate.

Force is defined as the product of the mass of the object and acceleration. The SI unit of the force of Newton.

Gravitational force is the force applied by gravity on an object. The gravity of the earth is 9.8 m/s².

Force = Mass × Acceleration

F = m × a

Mass of the crate = 10 kg

The initial speed of the crate = 1.50 m/s

The pulling force = 100 N

Horizontal angle made by the crate = 20 °

Coefficient of kinetic friction = 0.400

The crate is pulled to a height of 5 m.

$Sin θ= \frac{height}{distance}$

$= \frac{h}{d}$

$h = dSin θ$

Work done by the gravitational force on the crate is,

Work done by gravitational force = mass × gravity × height

$W_{g} = mgh$

$W_{g} = m \times g \times d \times sinθ$

$= 10 \times 9.8 \times 5 \times sin20°$

$= 167.67 \: J$

Therefore, 167.67 Joules of work is done by the gravitational force on the crate.

To know more about force, refer to the below link:

brainly.com/question/13191643

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1 year ago

the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar

adell [148]

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density)(volume)g

weight of object = (19.3)(0.55)g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

8 0

3 years ago

A pole vaulter takes 1.3 seconds to fall from peak height to the landing mat. What is his vertical velocity at contact with the

julia-pushkina [17]

$12.75 \frac{m}{s}$

Assuming vertical acceleration of $9.81 \frac{m}{s^2}$ , the speed after x seconds of falling is $9.81x \to 9.81(1.3) = 12.75$

6 0

2 years ago

Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc

Musya8 [376]

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

3 0

2 years ago

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

4 0

3 years ago