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ki77a [65]
2 years ago
6

A car travels at 40 mph. what distance will it travel in 3 hours?​

Physics
2 answers:
lukranit [14]2 years ago
8 0

Car speed = 40m/ hour

3 hours = 40 × 3 = 120m /hour

Margarita [4]2 years ago
7 0

Answer:

120

Explanation:

if the car travels 40 in one hour , to find the speed in 3 hours multiply 40 x 3 which will give us 120

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6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove
Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0
1 year 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
2 years ago
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
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