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Alex777 [14]
3 years ago
13

A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling

at 750 m/s. a) How much work was done by the rocket? b) What is the magnitude of the acceleration of the rocket? c) How long did the flight take?
Physics
1 answer:
hodyreva [135]3 years ago
4 0

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

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<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
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Answer:

Moment of inertia of the solid sphere:

I

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Is=25MR2...........(1)

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A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
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Answer:

a)

1.35 kg

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2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

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