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Tatiana [17]
3 years ago
9

Mr. Stephenson drives his car from Dallas to Town X, a distance of 437 km. The trip takes 7.27 hours. What is the average speed

of this trip in meters per second (m/s)?
Physics
2 answers:
Rom4ik [11]3 years ago
7 0
I believe the answer is <span>216396.148556 meters per second</span>
PtichkaEL [24]3 years ago
6 0

the answer is 60.1 m/s

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Which single force acts on an object in free fall
ICE Princess25 [194]

Answer:

gravity

Explanation:

4 0
3 years ago
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Sallys physical education teacher timed her run and recorded the time and distance in the table below. What is her average speed
Fudgin [204]
Answer: B) 2.5 m/s

Explanation: Find the average of the time and distance, and see how far they go in only 1 second.


1 + 2 + 3 + 4 + 5 = 15
15 divided by 5 = 3

3 seconds

2 + 5 + 7 + 10 + 12 = 36
36 divided by 5 = 7.2


7.2m per 3 seconds.

7.2 divided by 3 = 2.4

Therefore, the answer is technically 2.4m/s
4 0
4 years ago
To determine a waves' frequency, you must know the
givi [52]
The correct answer is<span> number of oscillations in a given period of time

This is measured in what is called the Hertz measurement and the period of time is usually taken to be per second.</span>
6 0
3 years ago
flipper (the dolphin) is out in the open ocean hunting tuna avec. he emits his pulse at 22khz and .42 seconds later he hears it
marusya05 [52]
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
v=\lambda f
where
v is the wave speed
\lambda its wavelength
f its frequency
Using \lambda = 2 cm = 0.02 m and f=22 kHz = 22000 Hz, we get
v=(0.02 m)(22000 Hz)=440 m/s

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
v= \frac{S}{t}= \frac{2L}{t}
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
L= \frac{vt}{2}= \frac{(440 m/s)(0.42 s)}{2}=92.4 m
5 0
3 years ago
A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach
Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

3 0
2 years ago
Read 2 more answers
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