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Ostrovityanka [42]
3 years ago
14

James decides to walk home from school today. He lives 3 miles from school and can walk home in 45 minutes. At what rate is Jame

s traveling?
A) 1 mile/10 minutes
B) 1 mile/15 minutes
C) 1 mile/20 minutes
D) 1 mile/30 minutes
Physics
1 answer:
Kaylis [27]3 years ago
3 0
B 1 mile/15 minutes is the right one
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Which of the following is not a means to accelerating?
GenaCL600 [577]
The answer is B. Remain still.
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3 years ago
What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?
sergey [27]

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

Learn more about acceleration at: brainly.com/question/12550364

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6 0
2 years ago
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
il63 [147K]

Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

cm. What is the length of the book in mm?

25. Explain the modifications

3 0
3 years ago
The humber bridge in england has the world's longest single span, 1410 m . calculate the change in length of the steel deck of t
Pepsi [2]
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length

Therefore,
ΔL = 12*10^-6*(18.5-(-3))*1410 = 0.36378 m
3 0
3 years ago
A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle i
olga55 [171]

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

6 0
3 years ago
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