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kolbaska11 [484]
3 years ago
14

A car is making a 40 mi trip. It travels the first half of the total distance 20.0 mi at 18.00 mph and the last half of the tota

l distance 20.0 mi at 56.00 mph. What is the car’s average speed in mph for the entire second trip?
Physics
1 answer:
sukhopar [10]3 years ago
5 0

Answer: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

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Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.
erica [24]
Wave D has the same frequency
5 0
3 years ago
One student bangs two bricks together
Romashka-Z-Leto [24]

Answer:

This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time

Explanation:

I hope this is what you need

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7 0
3 years ago
An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th
skelet666 [1.2K]

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

7 0
2 years ago
A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.
AlexFokin [52]

<h2><u>We have</u>,</h2>

  • Initial velocity (u) = 0 m/s
  • Time taken (t) = 2.9s
  • Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

  • Final velocity (v)
  • Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s \qquad … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m \qquad … ( Ans )

4 0
2 years ago
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