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victus00 [196]
3 years ago
6

The space shuttle fleet was designed with two booster stages. if the first stage provides a thrust of 53 kilo-newtons and the sp

ace shuttle has an acceleration of 18,000 miles per hour squared, what is the mass of the spacecraft in units of pounds-mass?
Physics
1 answer:
lidiya [134]3 years ago
4 0

To solve this problem we will apply Newton's second law, which indicates that the force is equivalent to the product between mass and acceleration, so

F = ma

Here,

F= Force

m = Mass

a = Acceleration

Rearranging to find the mass we have,

m = \frac{F}{a}

The value of the acceleration is

a = 18miles/hour^2 (\frac{0.00012417m/s^2}{1 miles/hour^2})

a = 0.002235m/s^2

Replacing to find the mass,

m = \frac{53kN}{0.002235m/s^2}

m = \frac{53*10^{-3}}{0.002235}

m = 23.71kg

Now in ponds this value is

m = 23.71kg(\frac{2.205lb}{1kg})

m=52.8 lb

Therefore the mass of the spacecraft is 52.8lb

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Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with
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Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

K=\frac{1}{2}Iw^2

Where I is rotational of inertia of solid cylinder about its central axis.

So

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2

Substitute the given values

So

K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

For Part (b)

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

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