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sashaice [31]
3 years ago
15

A bus travels 280 km south along a straight path with an average velocity of 88 km/h to the south. The bus stops for 24 min, the

n it travels 210 km south with an average velocity of 75 km/h to the south. a. How long does the total trip last? b. What is the average velocity for the total trip?
Physics
2 answers:
loris [4]3 years ago
8 0

Answer:

yes yes yes yes

Explanation:

because

IgorC [24]3 years ago
3 0

Answer:

76.78 km/h To calculate the average velocity for the total trip, you need to first determine the total distance traveled and the total time taken. First, let's calculate the total distance traveled. The trip consists of 2 legs. The 1st leg is 280 km and the 2nd leg is 210 km. So the total distance is 280 km + 210 km = 490 km. Now you need to calculate the total time taken. For this problem, there are 3 intervals that need to be accounted for. The travel time for the 1st leg, the duration of the rest stop in the middle, and the travel time for the 2nd leg. The travel time for both legs is calculated by dividing the distance traveled by the average speed. So for the first leg we have 280 km / (88 km / h) = 3.181818 h The 2nd leg is 210 km / (75 km/h) = 2.8 h The rest stop in hours is 24 min / (60 min/h) = 0.4 h The total time is 3.181818 h + 2.8 h + 0.4 h = 6.381818 h The average velocity is the distance divided by the time, giving: 490 km / (6.381818 h) = 76.78 km/h

Explanation:

Hope this helps!!

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Read 2 more answers
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0
2 years ago
Assignments
sladkih [1.3K]

Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

w=-F_f(d)

w=-(35)(20)

w=-700J

B) What is the work done on the cart by the gravitational force?  

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

w=Fdcos90

w=0

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J

D) Find the force the shopper exerts, using energy considerations.

F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N

E) What is the total work done?

You just need to add them:

w=wshopper+wfriction\\w=0

4 0
3 years ago
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