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3 years ago

Describe the effect of speed on voltage

Physics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

Speed has no effect on voltage.

Explanation:

The voltage of the battery in your car is always 12 to 13.5 volts. It makes no difference whether the car is home in the garage or zooming down the interstate.

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Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.

Ulleksa [173]

Answer:

If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.

The size of the circle determines.
The diameter should not be fixed either.

3 0

3 years ago

A sinusoidal wave of wavelength 2.00m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. At t = 0

Shkiper50 [21]

The frequency and angular frequency are 0.500 Hertz and 3.142 rad/s. respectively.
The angular wave number is equal to 3.142 rad/m.
The wave function for this wave is given by: y = Asin(kx - ωt + Φ).
The equation of motion for the left end of the string is given by: y = 0.100sin(3.142x - 3.142t + 0).
The equation of motion for the left end of the string at x = 1.50 m to the right is equal to y = 0.100sin(4.71 - 3.142t + 0).The maximum speed of any point on the string is 0.3142 m/s.

<h3>How to calculate the frequency and angular frequency?</h3>

First of all, we would determine the frequency of this wave by using this formula:

Frequency = wavelength/speed

Frequency = 0.100/2.00

Frequency = 0.500 Hertz.

For the angular frequency, we have:

Angular frequency, ω = 2πf

Angular frequency, ω = 2 × 3.142 × 0.500

Angular frequency, ω = 3.142 rad/s.

<h3>How to determine the angular wave number?</h3>

Angular wave number, k = 2π/∧

Angular wave number, k = (2 × 3.142)/2.00

Angular wave number, k = 3.142 rad/m.

<h3>How to determine the wave function for this wave?</h3>

Mathematically, the wave function for this wave is given by:

y = Asin(kx - ωt + Φ)

For the equation of motion for the left end of the string, we have:

y = 0.100sin(3.142x - 3.142t + 0)

For the equation of motion for the left end of the string at x = 1.50 m to the right, we have:

y = 0.100sin(3.142x - 3.142t + 0)

y = 0.100sin(3.142(1.5) - 3.142t + 0)

y = 0.100sin(4.71 - 3.142t + 0)

<h3>What is the maximum speed of any point on the string?</h3>

Vy = 0.100sin(- 3.142)cos(3.142x - 3.142t)

Vy ≤ 0.3142 m/s (since cosine varies +1 and -1).

Read more on wave function here: brainly.com/question/11181093

#SPJ4

Complete Question:

A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. Initially, the left end of the string is at the origin. Find:

(a) the frequency and angular frequency,

(b) the angular wave number, and

Determine the equation of motion in SI units for

(d) the left end of the string, and

(e) the point on the string at x = 1.50 m to the right of the left end.

(f) What is the maximum speed of any point on the string?

5 0

2 years ago

A box is 4 m in the air and has 120 J of potential energy. What is the box's mass

hodyreva [135]

Answer:

Mass of the box is 3.06 kg

Explanation:

Potential energy (that depends on gravity) is the type of energy which is stored that relies on the general position of different parts of a body. It is a property of a framework and not of an individual body or molecule; the framework made out of the Earth and the raised ball.

P.E = m.g.h Equation(1)

as we have to find the mass of the box then the formula for the mass will become

m=P.E/g(h) Equation(2)

as we know

P.E= 120J

h=4m

g=9.8m/s

g is the gravitational acceleration

putting values in the equation(2)

m = 120/9.8(4)

= 120/39.2

= 3.06 kg

mass of the box is 3.06 kg

5 0

4 years ago

If the length of a ramp (an inclined plane) is 12 ft, and it rises 2 ft, what is its MA?

dangina [55]

The MA is 6! Hope This Helps!

3 0

3 years ago

The most energy-efficient way to send a spacecraft to the Moon is to boost its speed while it is in circular orbit about the Ear

jeyben [28]

Answer:

98.33 %

Explanation:

On an elliptical orbit, angular momentum will be conserved .

Angular momentum = I ω = mvR

So mv₁R₁ = mv₂R₂

= v₁R₁ = v₂R₂

where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).

Here R₁ = Radius of the earth , R₂ is distance between moon and earth.

R₁ / R₂ = 1/60

v₁ /v₂ = R₂ / R₁ = 60

v₂ / v₁ = 1 / 60

1 - (v₂ / v₁ ) = 1 -( 1 / 60)

(v₁ -v₂)/v₁ = ( 60-1 )/60

(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %

5 0

4 years ago