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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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swat32

Answer:

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Dolphins sleep with one eye open

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A crocodile cannot stick its tongue out

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High Heels Were Originally Men's Shoes

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Explanation:

3 0
2 years ago
Read 2 more answers
What are successfulness of the Competition policy in South Africa?​
Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

hope this helps you

6 0
3 years ago
The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
3 years ago
Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?
lubasha [3.4K]
1) The equivalent resistance of two resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}
so in our problem we have
\frac{1}{R_{eq}} =  \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}
and the equivalent resistance is
R_{eq} =  \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A
4 0
3 years ago
Read 2 more answers
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
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