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3 years ago

How can a heavy moving van have the same momentum as a small motorcycle?

1 answer:

vladimir1956 [14]3 years ago

3 0

Hello there!

The formula for momentum is $p=mv$ .
p is momentum, m is mass, and v is velocity.

The mass of a heavy moving van will be much greater than the mass of a motorcycle, so how can they have the same momentum?

They can have the same momentum if the velocity of the moving van is less than the velocity of the motorcycle, or if the velocity of the motorcycle is greater than the moving van (same thing but differently worded)

Let the mass of the moving van be greater than the mass of the motorcycle by a factor of k. In order for the two to have the same momentum, the velocity of the motorcycle must be greater than the velocity of the moving van by the same factor, k.

Hope this helps! :)

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Acceleration and Force

olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0

2 years ago

12. An organ pipe that is 1.75 m long and open at both ends produces sound of

podryga [215]

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

8 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$