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Anestetic [448]
2 years ago
5

A student launches a small rocket which starts from rest at ground level. At a height of h = 1.25 km the rocket reaches a speed

of vf = 375 m/s. At that height the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward in this problem.
Physics
1 answer:
Novay_Z [31]2 years ago
4 0

The acceleration of the rocket will be "56.2 m/s²".

According to the question,

The initial speed during launch,

  • u = 0 m/s

The speed at fuel running out point,

  • v_f = 375 m/s

Height,

  • h = 1.25 km

           = 1250 m

As we know,

→ vf^2 = u^2+2ah

or,

→    a = \frac{v_f^2-u^2}{2h}

By putting the values, we get

→       =\frac{(375)^2-(0)^2}{2\times 1250}

→       =\frac{140625}{2500}

→       = 56.2  \ m/s^2

Thus the above solution is correct.

Learn more:

brainly.com/question/11038449

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Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
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Answer:

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3 years ago
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Straight wire of indefinite length (transient) passed by an electric current of 5.0 A. The magnetic field generated around this
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Answer:

C. 2.0 cm

Explanation:

The magnetic field around the wire at point M is given by Biot-Savart Law:

B = \frac{\mu_o I}{2\pi R}

where,

B = Magnetic field = 50 μT = 5 x 10⁻⁵ T

I = current = 5 A

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

R = distance of point M from wire = ?

Therefore,

5\ x\ 10^{-5}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(5\ A)}{2\pi R}\\\\R = \frac{(2\ x\ 10^{-7}\ N/A^2)(5\ A)}{5\ x\ 10^{-5}\ T}\\

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Answer:

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