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GarryVolchara [31]
2 years ago
7

Aluminum allows for the flow of electrons, while glass does not. Would an aluminum wire surrounded by glass be an effective desi

gn for a device that will need to carry an electric current?(1 point)
No, it would not. Both aluminum and glass are insulators, so the device would not carry a current.
Yes, it would. The glass would carry the current, while aluminum is an insulator.

Yes, it would. Aluminum is a conductor, and the glass would prevent electricity from transferring to a person holding the device

No, it would not. Both aluminum and glass are conductors, so the current would be exposed to outside elements.
Physics
1 answer:
pantera1 [17]2 years ago
3 0

answer

no

Explanation:

I do not think that I would because even though its a conductor in the insulator I think it would insulate it before it will work (not sure if that makes sense)

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A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general
valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength \lambda_i

• The wave does not satisfy the boundary conditions y_i(0;t)=0


Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions,y(0,1)=0


and y(L,t)=0

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength \lambda_i  can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, y_i(0;t)=0 and y_i(L;t)=0, that are satisfied only for particular value of \lambda_i  .

Given below are the incorrect options about the wave in the string.

•  A_i must be chosen so that the wave fits exactly o the string.

• Any one of  A_i or \lambda_i  or f_i  can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies f_i and the wavelength \lambda_i  must be such that y_i(0;t) = y_i(L;t)=0


(C)

Expression for the wavelength of the various normal modes for a string is,

\lambda_n=\frac{2L}{n} (1)

When n=1 , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

\lambda_n=\frac{2L}{1}\\\\2L

When n=2 , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

\lambda_n=\frac{2L}{2}\\\\L

When n=3, this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

\lambda_n=\frac{2L}{3}

Therefore, the three longest wavelengths are 2L,L and \frac{2L}{3}.

(D)

Expression for the frequency of the various normal modes for a string is,

f_n=\frac{v}{\lambda_n}

For the case of frequency of the i^{th} normal mode the above equation becomes.

f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

Therefore, the frequency of i^{th} normal mode is  f_i=\frac{v}{\lambda_i}

.

6 0
2 years ago
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
To protect her new two-wheeler, Iroda Bike
goldfiish [28.3K]

Answer:

The length of chain she is allowed is 1.169 ft

Explanation:

The given parameters are;

The linear density of the chain = 0.83 lb/ft

The weight limit of the chain she wants = 1.4 lb

The formula for linear density = Weight/length

Therefore, in order to keep the chain below 1.4 lb, we have;

Linear density = Weight/length

Therefore;

The maximum length she wants = Weight/(Linear density)

Which gives;

The maximum length she wants = 1.4 lb/(0.83 lb/ft) =1.169 ft

Therefore;

The length of chain she is allowed = 1.169 ft.

6 0
2 years ago
The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c
Natalka [10]

Answer:

EMF = 1684.67 Volts

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

EMF = \frac{d\phi}{dt}

now we have

\phi = BA

here since magnetic field is constant so we have

EMF = A\frac{dB}{dt}

now we have

A = (190 \times 10^3)(190 \times 10^3)

A = 3.61 \times 10^{10} m^2

now we have

EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})

EMF = 1684.67 Volts

6 0
2 years ago
1 point
e-lub [12.9K]
The engineer built a device called a generator
4 0
3 years ago
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