Which of the following describe an electric motor?

stepan [7]

D. Wind turbines, because wind is a renewable resource while all the other options use non renewable resources.

6 0

3 years ago

A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$

7 0

4 years ago

A race car starts from rest and travels east along a straight and level track. For the first 5.0 ss of the car's motion, the eas

irina [24]

Answer:

The acceleration of the car is 7.85 m/s²

Explanation:

Given;

vx(t)= (0.910m/s³)t², given time traveled by the car 't' = 5.0 s

⇒To determine the velocity for 5 seconds, we substitute in 5.0 s for t

vx(5)= (0.910m/s³)(5s)²

= (0.910m/s³)(25s²)

vx = 22.75 m/s

⇒To determine the acceleration of the car when vx=12.0m/s

Acceleration is change in velocity per unit time

when vx=12.0m/s, our new equation becomes; 12 = (0.910m/s³)t²

Solving for t: t² = 12/0.91

t² = 13.187

t = √13.187 = 3.63 s

Acceleration = Δv/Δt

$= \frac{(22.75 - 12)m/s}{(5-3.63)s} = (\frac{10.75}{1.37}).\frac{m}{s^2}$

Acceleration = 7.85 m/s²

3 0

4 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s.

Anton [14]

Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

leg 1: 24.5 m/s

leg 2: 49 m/s

leg 3: 23.9 m/s

Complete trip: 43.8 m/s

Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.

Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

x = 490 m

The velocity reached in that phase is:

v = v0 + a · t

v = 0 m/s + 2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s (without rounding the final position the result is 49.0 m/s)

Phase 3:

v = 5.32 × 10³ m - 5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v = 5.32 × 10³ m - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

7 0

4 years ago

A sound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3

Lina20 [59]

Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is
I = k/r^2
where
k = constant
r = radius

When r=1, the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
= 10 log₁₀ [(1/9)*(I₁/I₀)]
= 10 log₁₀(1/9) + 270
= 260.5

Answer: 260.5 dB (nearest tenth)

4 0

3 years ago