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valentinak56 [21]
2 years ago
12

Explain two scenarios where a large truck can have the same momentum as a small car.

Physics
1 answer:
KengaRu [80]2 years ago
5 0

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

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or
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3 years ago
Robert dropped his new iPhone from his balcony. It hit the ground 3.5 seconds later. What was the height of his balcony?
Lorico [155]

its B. 60 meters

Explanation:

cause I looked up a calculator and solved it

8 0
2 years ago
Can someone explain how they got their answer or how I get the change in number? :(
enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0
2 years ago
The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known
Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

m_1 = Mass of electron = 9.11\times 10^{−31}\ kg

v_1 = Speed of electron = 5.7\times 10^7\ m/s

p_2 = Neutrino has a momentum = 7.3\times 10^{-24}\ kg m/s

M = total mass = 2.34\times 10^{-26}\ kg

In the x axis as the momentum is conserved

Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s

In the y axis

Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s

The resultant velocity is

R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s

The recoil speed of the nucleus is 2240.92365 m/s

3 0
3 years ago
A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
2 years ago
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