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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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A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more
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3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

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3200÷0.22= 145.4545...N
(it is an infinite decimal)
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3 years ago
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