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ikadub [295]
3 years ago
6

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the

same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse
Physics
2 answers:
WARRIOR [948]3 years ago
5 0

Answer:

The head on Collison because you have both cars going (for the sake of it 30mph) and they both collide the energy from that is 60mph because the speed is combined with the two cars.

Marat540 [252]3 years ago
4 0

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

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Which circuit generates the most power
Tom [10]

Answer:

parallel circuit

Explanation:

In a parallel circuit, the potential difference across each of the resistors that make up the circuit is the same. This leads to a higher current flowing through each resistor and subsequently the total current flowing through all the resistors is higher.

5 0
3 years ago
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An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys
sammy [17]

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

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7 0
3 years ago
Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k
Harrizon [31]

Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

Mass of the electron, m_e=9.11\times 10^{-31}\ kg

Mass of the proton, m_p=1.67\times 10^{-27}\ kg

The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

\lambda_e=1.09\times 10^{-10}\ m

The wavelength of the proton is given by :

\lambda_p=\dfrac{h}{m_p v}

\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}

\lambda_p=5.96\times 10^{-14}\ m

(b) Kinetic energy, K=1.71\times 10^{-15}\ J

The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}

\lambda_p=2.77\times 10^{-13}\ m

Hence, this is the required solution.

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