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natulia [17]
3 years ago
10

A massive tractor rolls down a country road. in a perfectly inelastic collision, a small sports car runs into the machine from b

ehind. (i) which vehicle experiences a change in momentum of larger magnitude
Physics
2 answers:
Helga [31]3 years ago
5 0
The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change. 
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
Tresset [83]3 years ago
3 0

Answer:

c) Their momentum changes are the same size.

Explanation:

In order to understand this you need to remember that the momentum is divided equally in the collision, so the correct answer to this is the fact that their momentum will changes the same, because that is how energy is preserved, the thing that could be different is that they are affected different by that momentum, it would probably affect differently the small car and the tractor, it would probably have a lot of impact for the small car.

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Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th
alukav5142 [94]

Answer:

The altitude of geostationary satellite is 3.58\times10^{7}\ m

Explanation:

Given that,

Radius of moon's orbit r=3.84\times10^{8}\ m

Time period T=2.36\times10^{6}\ sec

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

T=\sqrt{\dfrac{4\pi^2}{GM}a^3}

a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}

a=4.217\times10^{7}\ m

We need to calculate the altitude of geostationary satellite

Using formula of altitude

h = a-R_{e}

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

h =4.217\times10^{7}-6.38\times10^{6}

h =35790000\ m

h=3.58\times10^{7}\ m

Hence, The altitude of geostationary satellite is 3.58\times10^{7}\ m

4 0
3 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

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3 years ago
Which of the following best describes a coastal plain?
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3 years ago
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Answer is 3. Volume= mass divided by density
4 0
3 years ago
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Solved, can someone check it over! ​
kakasveta [241]

Answer:

its good no need to change anything :))

4 0
3 years ago
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