PHYSICS HIGH SCHOOL

50 cm to mA flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω, what is the magnitude of the induced current?

Answers

Answer 1

Answer:

Answer:

0.025 A

Explanation:

A = 50 cm^2 = 50 x 10^-4 m^2

B2 = 6 T, B1 = 2 T

db = 6 - 2 = 4 T

dt = 2 s

R = 0.4 ohm

Let i be the magnitude of induced current and e be the induced emf.

According to the Faraday's law of electromagnetic induction

e = dФ / dt

e = A dB / dt

e = 50 x 10^-4 x 4 / 2 = 0.01 V

i = e / R = 0.01 / 0.4 = 0.025 A

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I will mark brainliest if you get it right.

A boy is running with kinetic energy of 810j if the boy has a mass of 80kg what is his peed

Answers

KE=(1/2)mv² ⇒ v=√(2KE/m)

Data:
KE=kenetic energy=810 J
m=mass=80 Kg
v=velocity (speed)

v=√(2KE/m)
v=√(2*810 J/80 Kg)=4.5 m/s.

answer: his speed is 4.5 m/s.

In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answers

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$ .

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$ .

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-(1)/(2)g{t^2}$ ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=(1)/(2)g{t^2}$

Rearrange the above expression for $t$ .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (2)

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \n&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \n \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \n&=1.116\,{\text{s}} \n \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$ .

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$ .

$\boxed{v=(x)/(t)}$ …… (3)

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \n&=19.71\,{\text{ft/s}} \n \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Learn more:

1. Energy density stored in capacitor brainly.com/question/9617400

2. Kinetic energy of the electrons brainly.com/question/9059731

3. Force applied by the car on truck brainly.com/question/2235246

Keywords:

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Final answer:

Using the principles of projectile motion from Physics, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to reach the far side of the river.

Explanation:

This problem can be solved using basic Physics, specifically projectile motion. Here, Captain Sam Brady had to run off the edge of the cliff to make it safely to the far side of the river which is 22 ft away while falling 20 ft down. We assume that he jumps horizontally (i.e., his initial vertical velocity is 0).

Firstly, we calculate the time for the vertical fall. Using the equation t = sqrt (2h/g) where h is height and g is the acceleration due to gravity (32.2 ft/s²), we get time t ≈ 1.12s (rounded to two significant figures).

Next, we can use this time to figure out his initial horizontal velocity needed. The equation v = d/t where v is velocity, d is distance, and t is time gives us v ≈ 19.64 ft/s (rounded to two significant figures).

So, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to make it safely across the river.

Learn more about Projectile Motion here:

brainly.com/question/20627626

#SPJ2

An engine raises 100kg of water through a height of 60m in 20s. what is the power of the engine

Answers

Answer:

the answer is 3000

Explanation:

power=mass×gravity×height or (divided by /Time)

so 100kg×10×60m/20s

=60,000/20

=3000joule

if I am wrong correct me

Which will have more resistance: 220V, 100W bulb or a 220V, 60W bulb?

Answers

The power dissipated by something with voltage across it is

(voltage)² / (resistance) .

You can see that resistance is in the denominator of the fraction,
so larger resistances dissipate less power, and smaller resistances
dissipate more power.

Here's an easy way to remember it:
-- Connecting a piece of rubber (high resistance) across the terminals of a battery
uses very little power from the battery.
-- Laying a piece of wire (low resistance) across the terminals of a battery
may draw so much power from the battery that the wire melts.

Since the voltage across both bulbs is expected to be the same, the bulb
designed to dissipate 60 watts has higher resistance than the one designed
to dissipate 100 watts.

What is the weight, on Earth, of a book with a mass of 1.5 kg