Which of the following best describes what occurs in a fusion reaction?

A segment A of wire stretched tightly between two posts a distance L apart vibrates in its fundamental mode with frequency f. A

____ [38]

Answer:

option (c)

Explanation:

Fundamental frequency of segment A = f

Second harmonic frequency of B = fundamental frequency of A .

Tension in both the wires is same and the mass density is also same as the wires are identical.

fundamental frequency of wire A is given by

$f=\frac{1}{2L_{A}}{\sqrt{\frac{T}{m}}}$ .... (1)

Second harmonic of B is given by

$f=\frac{2}{2L_{B}}{\sqrt{\frac{T}{m}}}$ .... (2)

Equation (1) is equal to equation (2), we get

$\frac{1}{2L_{A}}=\frac{2}{2L_{B}}$

$L_{B}=2L_{A}$

So, LB = 2 L

Thus, the length of wire segment B is 2 times the length of wire segment A.

3 0

3 years ago

A wire carries a current of 10 amps in a direction of 90 degrees with respect to the direction of an external magnetic field of

Oksi-84 [34.3K]

Answer:

15 N

Explanation:

The magnetic force on a piece of current-carrying wire is given by:

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the piece of wire

B is the magnetic field strength

$\theta$ is the angle between the direction of B and I

In this problem:

I = 10 A

B = 0.3 T

L = 5 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$F=(10 A)(0.3 T)(5 m) sin 90^{\circ}=15 N$

6 0

3 years ago

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→ θ ≈ 3.52°

8 0

3 years ago

If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intens

ankoles [38]

The force of a charge in an electric field is:

$\vec{F}_e=q\vec{E}$

In this case we know the electric field is:

$\vec{E}=104\hat{j}$

and that the charge is that of the electron, then we have:

$\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}$

Therefore, the magnitude of the force is

$1.664\times10^{-17}\text{ N}$

and in points down.

The weight of the electron is:

$\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}$

Making the quotient between the force we have:

$\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9$

Therefore, the electric force is approximately 1e9 times the weight.

3 0

1 year ago

You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo

IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

= (0.81) (1.42) (μ m g)

= (0.81) (1.42) (200)

= 230 N

4 0

3 years ago

Which of the following best describes what occurs in a fusion reaction?

Which of the following best describes what occurs in a fusion reaction?