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3 years ago

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A car is stopped at a red light. When the light changes, the car begins to accelerate at a rate of 10 m/s^2 and it continues to

accelerate over the next 25 seconds. What is the velocity of the car after after it accelerates for 25 s?

1 answer:

yaroslaw [1]3 years ago

4 0

The velocity of the car after it accelerates for 25 seconds would be 820.2 ft/s or 250 m/s

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If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in

Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, $N_P$ = 50 turns

number of turns in the secondary winding, $N_S$ = 10 turns

the secondary load resistance, $R_S$ = 250 Ω

Determine the turns ratio;

$K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5$

Now, determine the reflected resistance in the primary winding;

$\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms$

Therefore, the reflected resistance in the primary winding is 6250 Ω

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3 years ago

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rjkz [21]

Answer:

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2 years ago

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

as we know that

$m_1 = 5 kg$

$m_2 = 2 kg$

$v_{1i} = 2.8 m/s$

$v_{2i} = 0 m/s$

now from above equation we have

$5(2.8) + 2(0) = m_1v+ m_2v$

$14 = (5+ 2) v$

$v = 2 m/s$

so the speed of combined system is 2 m/s

8 0

3 years ago

A race car accelerates from 0 m/s to 30.0 m/s with a displacement of

worty [1.4K]

Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

$1. \ a=\frac{v-v_0}{t};$

$2. \ L=\frac{at^2}{2}.$

II) using these two equations after substitution v₀=0; v=30 and L=45:

$\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.$

$\left \{ {{at^2=90} \atop {at=30}} \right. \ \ \left \{ {{a=10} \atop {t=3}} \right.$ $=> \ a=10\frac{m}{s^2}$

6 0

3 years ago

Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of

alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

$\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx 87.084 \ ft./s$

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u /> $v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h$ <u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

$v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}$

$v = \dfrac{40,000 \times 30 + 150 \times 0}{40,000 + 150} \approx 29.89$

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0

3 years ago