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3 years ago

A traveler covers a distance of 413 miles in a time of 7 hours. What is the average speed for his trip?

Physics

1 answer:

Nina [5.8K]3 years ago

7 0

Answer:

59/mph.

Explanation:

We just have to divide the number of miles, by the amount of time it took. 413 miles / 7 hours = 59 miles per hour.

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Answer:

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3 years ago

What is the momentum of a 10kg ball moving at a velocity of 4m/s east?

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Answer:

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Explanation:

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A spacecraft is traveling in interplanetary space at a constant velocity. What is the estimated distance traveled by the spacecr

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3 years ago

Read 2 more answers

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$