Lol gn sleep well:))

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

$\epsilon = - d\frac{\phi_B}{dt}$

$\epsilon = -NA\frac{dB}{dt}$

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

$\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}$

$\epsilon = 0.212V$

$\epsilon = 0.21V$

Therefore the correct answer is A.

6 0

3 years ago

A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b

netineya [11]

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

h^2 = 25 = 100

h Sqrt(75)

h = 8.66 ft

Put h = 8.66ft in eq2

2 × 8.66 × (-2) + 2 ×5 dl/dt

dl/dt = 17.32 / 5

dl/dt = 3.464ft/sec

7 0

2 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

3 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>:

$F=k (x-x_{o})$ (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$ (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$ is the gravity acceleration.

$W=(0.25kg)(9.8\frac{m}{s^{2}})$ (3)

$W=2.45N$ (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$ , we can substitute the values in equation (1) and find $k$ :

$W=k (x-x_{o})$ (5)

$2.45N=k (0.18m-0m)$ (6)

<u>Finally:</u>

$k=13.61\frac{N}{m}$

4 0

2 years ago

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

Nataliya [291]

Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0

2 years ago