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timama [110]
3 years ago
11

The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 ​kilo-newtons and the

space shuttle has an acceleration of 16,000 miles per hour squared ​, what is the mass of the spacecraft in units of​ pounds-mass ​?
Physics
1 answer:
Tresset [83]3 years ago
8 0

Answer:

m = 81281.5 pounds.

Explanation:

Given that,

Force, F = 73 kN

Acceleration of the space shuttle, a = 16000 mi/h²

1 miles/h² = 0.0001241 m/s2

16000 mi/h² = 1.98 m/s²

We need to find the mass of the spacecraft.

According to Newton's second law,

F = ma

m is mass of the spacecraft

m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg

Since, 1 kg = 2.20462 pounds

m = 81281.5 pounds

Hence, the mass of the spacecraft is 81281.5 pounds.

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A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal
Mkey [24]

Answer:

   F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

     I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

      cos 45 = vₓ / v

      vₓ = v cos 45

      vₓ = 8 cos 45

      vₓ = 5,657 m / s

We write the moment

Before the crash                          p₀ = m vₓ

After the shock                            p_{f} = -m vₓ

The variation of the moment      Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis           Fₓ t = Δp

       

        Fₓ = 2 m vₓ / t

        Fx = 2 0.450 5.657 / 0.250

        Fx = 20.4 N

We perform the same calculation on the y axis

       sin  45 = vy / v

       vy = v sin 45

       vy = 8 sin 45

       vy = 5,657 m / s

We calculate the initial momentum   po = m v_{y}

Final moment                                      p_{f} = m v_{y}

Variations moment                             Δp = mv_{y} - mv_{y} = 0

Force in the Y-axis                             F_{y} = 0

Therefore the total force is

       F = fx i ^ + Fyj ^

       F = Fx i ^

       F = 20.4 i ^

3 0
3 years ago
In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. An
evablogger [386]

According to the plot, the positions at time <em>t</em> = 0 s and <em>t</em> = 19 s are -1 m and -2 m, respectively. So the average velocity for the 19-s interval is

v_{\rm ave} = \dfrac{-2\,\mathrm m - (-1\,\mathrm m)}{19\,\mathrm s} = -\dfrac1{19}\dfrac{\rm m}{\rm s}\approx \boxed{-0.05\dfrac{\rm m}{\rm s}}

8 0
3 years ago
An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac
Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

       = (50 - 0)/4

       = 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0
3 years ago
One of Lex Luthor's henchman attacks Superman, shooting a rapid-fire stream of 3.3 g bullets at him at a rate of 112/min. The sp
igor_vitrenko [27]

Answer:

3.2451N

Explanation:

Mass of the bullet (m) = 3.3g = 3.3*10^{-3}Kg

Speed of the bullet (V)= 527m/s

Rate of bullet (r) = 112/min = 1.866\sec

We can calculate with this information the average acceleration of bullets

a=V*r = 527\frac{m}{s}\frac{1.866}{s} = 983.38m/s^2

The force is given by,

F=ma\\F=(3.3*10^{-3})*983.38m/s^2 = 3.2451N

That is just because he is Superman.

4 0
3 years ago
A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s
Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

             p₀ = m v₀

celestines que clash case

             p_f = (m + M) v

             po = pf

             m v₀ = (n + M) v

             v = \frac{m}{m+M}

calculemos

            v= \frac{0.200}{0.200+M} 0.450

            v= 0.09 m/s

elastic shock case

           p₀ = m v₀

           p_f = m v₁ +M v₂

           p₀ = p_f

           m v₀ = m v₁ + m v₂

6 0
3 years ago
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