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Lady_Fox [76]
3 years ago
8

The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I

t weighs 37,500 pounds.) If you mounted this aircraft engine on your car, what acceleration would you get? (Please use metric units. The data in pounds are given for comparison. Use a reasonable estimate for the mass of your car. A kilogram mass weighs 2.2 pounds.)
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

The acceleration you can get with that engine in your car is around 70,56 (\frac{m}{s^{2} }) or 7,26 (\frac{ft}{s^{2} } ) using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

F=m*a

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }

a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }

<em> Note: N or Newton units are: \frac{kg * m}{s^{2} }</em>

a= 70,54 \frac{m}{s^{2} }

Also in pounds you can compared

a= \frac{2400  lf }{3 306  lf}

Note: lf in force units are: \frac{lf*ft}{s^{2} }

a=7,26 \frac{ft}{s^{2} }

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Answer:

915 Hz

Explanation:

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f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

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f₀ = 983 [(343+28.4)/(343+55.9)]

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Answer:

The <u>nucleus</u> is the core of an atom.

Step-by-step explanation:

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5 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

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5 0
4 years ago
A) the unstretched length of each elastic rope is 24m. The rope obeys hookes law. The vertical distance between P and Q is 35m.
solong [7]

Explanation:

a) The rope obeys Hooke's law, so:

F = k Δx

The elastic energy in the rope is:

EE = ½ k Δx²

Or, in terms of F:

EE = ½ F Δx

Use trigonometry to find the stretched length.

cos 20° = 35 / x

x =  37.25

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Δx = 13.25

The elastic energy per rope is:

EE = ½ (3.7×10⁴ N) (13.25 m)

EE = 245,000 J

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b) The elastic energy in the ropes is converted to gravitational energy.

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5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h

h = 42 m

Rounded to one significant figure, the height is 40 m.  So the claim is not justified.

6 0
3 years ago
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