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3 years ago

Ford used to make a 7.0 liter engine. Calculate how many many cubic inches would be the same size engine.

Physics

1 answer:

Anestetic [448]3 years ago

3 0

Answer : The volume of engine in cubic inches is, $427.166\text{ inch}^3$

Explanation :

As we are given that the volume of engine is, 7.0 liters. Now we have to calculate the volume of engine in cubic inches.

Conversion used :

$1\text{ liter}=1000cm^3$

So, $7\text{ liter}=7000cm^3$

$1\text{ inch}=2.54cm$

or,

$1cm=\frac{1}{2.54}\text{ inch}$

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

As,

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

So,

$7000cm^3=\frac{7000cm^3}{1cm^3}\times (\frac{1}{2.54}\text{ inch})^3=427.166\text{ inch}^3$

Thu, the volume of engine in cubic inches is, $427.166\text{ inch}^3$

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Increasing the mass attached to a spring will decrease the angular frequency of its vibrations. True or False

Scrat [10]

Answer:

decreases

Explanation:

The angular frequency of the mass is given by

$\omega =\sqrt{\frac{k}{m}}$

where, k is the spring constant and m be the mass of the body which is doing oscillations.

if the mass of the body increases, so the value of angular frequency decreases, as the angular frequency is inversely proportional to the square root of the mass of the body.

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3 years ago

2. Un ciclista que está en reposo comienza a pedalear hasta alcanzar los 16.6km/h en 6 minutos. Calcular la distancia total que

jek_recluse [69]

Answer:

$s = 5822.28\,m\,(5.822\,km)$

Explanation:

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$a = \frac{4.611\,\frac{m}{s}-0\,\frac{m}{s}}{360\,s}$

$a = 0.013\,\frac{m}{s^{2}}$

La distancia recorrida es (The travelled distance is):

$s = \frac{1}{2}\cdot \left(0.013\,\frac{m}{s^{2}} \right)\cdot (360\,s)^{2} + \left(4.611\,\frac{m}{s} \right)\cdot (1080\,s)$

$s = 5822.28\,m\,(5.822\,km)$

3 0

3 years ago

A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on

muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

$x=ucos\alpha t-(1/2)gt^{2}$ but the acceleration in the horizontal path is zero, hence we have

$x=ucos\alpha t$

Since the horizontal distance covered is 155m at 7.6secs, we have $ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1$

Also from the vertical path, the distance covered is expressed as

$y=usin\alpha t-(1/2)gt^{2}$

since the horizontal distance covered in 7.6secs is 195m, then we have

$y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2$

Hence if we divide both equation 1 and 2 we arrive at

$\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\$

Hence if we substitute the angle into the equation 1 we have

$ucos72.02=20.38\\u=66.02m/s$

Hence the initial velocity is 66.02m/s

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4 years ago

An alkane has molecular weight 86. Write its molecular formula. What will be its physical state?

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C6H14

Gaseous state

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3 years ago

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KonstantinChe [14]

I think the answer is D.
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4 years ago