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ElenaW [278]
3 years ago
10

A 4,000 kg truck is moving at +10 m/s hits a 1500 kg parked car which moves off at +10 m/s. What is the velocity of the truck

Physics
1 answer:
Igoryamba3 years ago
6 0

Answer:

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s

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The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

<u><em>L = 57.88 mm</em></u>

<u><em /></u>

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

Read more at; brainly.com/question/17161594

8 0
2 years ago
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Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

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N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

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