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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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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
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Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

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q_{1} is the heat absorbed by the solid at 0⁰C

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Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

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n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

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=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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