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4 years ago

12

Yosef applies an input force of 50 N to a crowbar. The crowbar applies a force of 750 N to the lid of a crate. What is the mecha

nical advantage of the crowbar? The mechanical advantage of the crowbar is .

2 answers:

docker41 [41]4 years ago

5 0

The mechanical advantage of the crowbar is 15.

Ymorist [56]4 years ago

5 0

The answer to your question is 15

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The scanner records the time from when a ultrasound wave is emitted to when its reflection is received

telo118 [61]

Explanation:

The scanner records the time from when a ultrasound wave is emitted to when its reflection is received. A technician calculates the depth of the reflection using the equation as :

$\text{depth}=\dfrac{1}{2}\times \text{speed of ultrasound in patient}\times \text{time recorded by scanner}$

The distance covered by ultrasonic wave when it was emitted and gets reflected is 2d. Speed is given by :

$v=\dfrac{d}{t}$

d is distance and t is time

Here, d = 2d

So, the factor of (1/2) is used because the distance covered by the wave is 2 times when it was emitted and received.

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3 years ago

Escape speed from the surface of earth is about

dezoksy [38]

<span>On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.</span>

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3 years ago

Read 2 more answers

A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the conta

Novay_Z [31]

The right answer is (b)

8 0

3 years ago

What’s the answer to this question

alexgriva [62]

Answer:

space = 66.24 [m]

Explanation:

To solve this problem we must remember that the average speed is defined as the relationship between a space traveled over a certain time.

$Av = \frac{space}{time}$

where:

space [m]

Av = average velocity = 3.6 [m/s]

time = 18.4 [s]

$space = 3.6*18.4\\space = 66.24 [m]$

7 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