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iogann1982 [59]
3 years ago
12

A 1000 Kg car traveling at 10 m/s hits the back of a 5000 Kg truck

Physics
1 answer:
insens350 [35]3 years ago
8 0

Answer:

0 m/s

The car becomes stationary

Explanation:

The law of conservation of linear momentum states that the sum of inital and final momentum should be equal and expressed as

m_cu_c+m_tu_t=m_cv_c+m_tv_t

Where m represent the mass, u and v are tge initial and final velocities while subscripts c and t represent car and truck.

Taking forward direction as positive then considering that the truck is originally at rest, we substitute original truck velocity with 0, mass of car and truck with 1000 kg and 5000 kg respectively then final truck velocity as 2 m/s as we take initial car velocity to be 10 m/s

1000*10+(5000*0)=5000*2+1000v

1000v=0

V=0

Therefore, the car finally becomes stationary.

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A rifle of mass 2 kg is horizontally suspended by a pair of strings so that recoil can be measured. The rifle fires a bullet of
aliya0001 [1]

Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

Vg = 1m/s

3 0
3 years ago
Given the displacement vector D = (4î − 8ĵ) m, find the displacement vector R (in m) so that D + R = −3Dĵ. (Express your answer
Elena-2011 [213]

Answer:

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Explanation:

4 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
An object has a mass of 2,000kg. what is its weight on earth? show your work​
artcher [175]

Answer:

F=m*g is the formula and the answer is 19,620 kg

Explanation:

Since the formula is F=m*g and Earth's gravity is 9.81 m/s^2 all you need to do is multiply 2,000 by 9.81

6 0
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