Pls help i’ll give brainliest if you give a correct answer!!

a 1.2 kg rocket is launched from ground level with initial velocity of 12 m/s. What is the maximum height the rocket can reach?

Tanzania [10]

As the rocket is launched from the ground its height will go on increasing till it stops

So the height of rocket will be maximum when its speed becomes zero

so here we can use energy conservation theory

$PE = KE$

$mgH = \frac{1}{2}mv^2$

$9.8*H = \frac{1}{2}*12^2$

$H = 7.35 m$

So it will reach upto height 7.35 m

5 0

3 years ago

A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it

Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

$I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}$

The required initial angular speed is obtained herein:

$g= \omega_{o}^{2}\cdot R_{ss}$

$\omega_{o}=\sqrt{\frac{g}{R_{ss}} }$

$\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }$

$\omega_{o} \approx 0.253\,\frac{rad}{s}$

The initial moment of inertia is:

$I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}$

The final moment of inertia is:

$I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}$

Now, the final angular speed is obtained:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )$

$\omega_{f} = 0.331\,\frac{rad}{s^}$

The apparent acceleration is:

$a_{f} = \omega_{f}^{2}\cdot R_{ss}$

$a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)$

$a_{f} = 16.763\,\frac{m}{s^{2}}$

This is approximately 1.709g.

4 0

4 years ago

how you could calculate a time step (dt) if you have a vector of accumulative distance (ie the distance keeps adding every step)

lakkis [162]

Answer:

dt = ds/dv.

Explanation:

to calculate the time step if you have a vector of accumulative distance and instanteneous velocity vector can be expressed bellow

solution

ds = dv/dt

divide the step distance by the instantaneous velocity you will get the time step.

that it

dt = ds/dv.

so to calculate the time step having a vector that keeps adding every step and speed at that moment is this dt = ds/dv.

6 0

4 years ago

Read 2 more answers

The toxic effects of water pollution are generally _______.

nikdorinn [45]

<span>Immediate and long lasting because they pollute the water itself which is the most important commodity for humans. Dirty water gives us all types of diseases which are all long lasting and the body has a immediate reaction to the pollutants in the water. Drinking polluted water or even bathing in it can be very harmful to the human body.</span>

7 0

3 years ago

A damping force affects the vibration of a spring so that the displacement of the spring is given by y = e−4t(cos 2t + 3 sin 2t)

astraxan [27]

Answer:

Average velocity = dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

Explanation:

Velocity is defined as the rate of change in displacement.

Velocity = change in displacement/time taken

Given the displacement of the string as;

y = e^−4t(cos 2t + 3 sin 2t).

To get the average velocity, we will find the derivative of the displacement with respect to time.

Using function of a function to solve this;

Let u = 4t(cos 2t + 3 sin 2t)... (1)

y = e^-u... (2)

Differentiating both functions with respect to their variables we have;

dy/du = -e^-u

du/dt is gotten using the product rule to have;

du/dt = 4t(-2sin2t+6cos2t)+4(cos2t+3sin2t)

Opening up the bracket we have;

du/dt = -8tsin2t+24tcos2t+4cos2t+12sin2t

Collecting like terms;

-8tsin2t+12sin2t+24tcos2t+4cos2t

du/dt = -4sin2t(2t-3)+4cos2t(6t+1)

dy/dt = dy/du × du/dt

dy/dt = -e^-u × -4sin2t(2t+3)+4cos2t(6t+1)

Substituting u = 4t(cos 2t + 3 sin 2t) into dy/dt, we will have;

dy/dt = -e^-4t(cos 2t + 3 sin 2t) × -4sin2t(2t+3)+4cos2t(6t+1)

dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

The average velocity of the wave function therefore give us;

dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

4 0

3 years ago