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3 years ago

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A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati

on if it takes off vertically. How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust? In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.

1 answer:

Kazeer [188]3 years ago

4 0

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2$

Initial acceleration of the rocket is 17.5 m/s²

$v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s$

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

$\Delta v=v_{e}\ln \frac{m_0}{m_f}$

where,

$m_0$ = Initial mass of rocket with fuel

$m_f$ = Final mass of rocket without fuel

$v_e$ = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

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Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

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Answer:

4. It is the force of the road on the tires (an external force) that stops the car.

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If there is no friction between the road and the tires, the car won't stop.

You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.

The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.

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Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.

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adell [148]

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

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m1= 470 kg

x= 4m

m2= 75kg

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The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

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τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

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