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snow_tiger [21]
3 years ago
8

A truck moves 60 kilometers east from point A to point B. At point B, it turns back west and stops 15 kilometers away from point

A. What are the total distance and total displacement of the truck? A. The total distance is 105 kilometers, and the total displacement is 15 kilometers east. B. The total distance is 105 kilometers, and the total displacement is 45 kilometers east. C. The total distance is 60 kilometers, and the total displacement is 60 kilometers east. D. The total distance is 60 kilometers, and the total displacement is 45 kilometers east. E. The total distance is 105 kilometers, and the total displacement is 75 kilometers east.
Physics
2 answers:
nata0808 [166]3 years ago
6 0

A truck moves 60 kilometers east from point A to point B. At point B, it turns back west and stops 15 kilometers away from point A.

Displacement=15km east

Distance=60+(60-15)=105km

ans is A. The total distance is 105 kilometers, and the total displacement is 15 kilometers east.

sineoko [7]3 years ago
3 0

Answer:

a

Explanation:

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A loudspeaker in a parked car is producing sound whose frequency is 20 510 Hz. A healthy young person with normal hearing is sta
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Answer:

The car must be moving away from the person.

Explanation:

From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.

The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.

<u>Due to this reason, the car must be moving away from the person.</u>

4 0
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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
1 year ago
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