Waves that make up the visible part of the electromagnetic spectrum have

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a

Naily [24]

Answer:

The angular acceleration of the pencil α = 17 rad·s⁻²

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

τ = I α (1)

W r = I α (2)

The weight is that the pencil has is,

sin 10 = r / (L/2)

r = L/2(sin(10))

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

I = 1/3 M L²

Thus,

mg(L / 2)sin(10) = (1/3 m L²)(α)

α(f) = 3/2(g) / Lsin(10)

α = 3/2(9.8) / 0.150sin(10)

α = 17 rad·s⁻²

Therefore, the angular acceleration of the pencil is 17 rad·s⁻²

3 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Describe 2 ways that mass and weight are the same

navik [9.2K]

Anything that has mass has weight and anything that has weight has mass simple.

7 0

3 years ago

A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is

ivanzaharov [21]

anwser will be

F = ma

where

F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet

Substituting appropriately,

F = 0.005a --- call this Equation 1

Next working equation is

Vf^2 - Vo^2 = 2as

where

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel

Substituting appropriately,

326^2 - 0 = 2(a)(0.83)

a = 64,022 m/sec^2

the anwser will be
Substituting this into Equation 1,

F = 0.005(64,022)

F =320.11 Newtons

Hope this helps.

8 0

3 years ago

Please help me on this l

eduard

Answer:

16

Explanation:

because

7 0

3 years ago