Ask question

Ask question

All categories

2 years ago

14

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]2 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

You might be interested in

You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa

rusak2 [61]

The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0

3 years ago

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0

2 years ago

What happens to the force between the spheres when you increase the mass of one of the spheres?

andrew11 [14]

The force between the spheres increases when the mass increases in one of the spheres.

<u>Explanation:</u>

Newton law of universal gravity extends gravity beyond the earth's surface. This gravity depends directly on the mass of both objects and is inversely proportional to square of distance between their centers.

$\bold{F=\frac{G \times\left(m_{1} \times m_{2}\right)}{\left(r^{2}\right)}}$

Since gravity is directly proportional to “mass of both interacting objects”, stronger objects with greater gravitational force attract. If the mass of one object increases, gravity between them also increases. For example, if an object's mass of one double, force between them also doubles.

4 0

2 years ago

Read 2 more answers

The left end of a rod of length and rotational inertia is attached to a frictionless horizontal surface by a frictionless pivot,

natima [27]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

6 0

2 years ago

How long does it take to change a flat plain into mountains?

ss7ja [257]

It would take millions of years to form a mountain as plates move very slowly and to form it first one plate should climb upon another. After this very slowly this hill will convert into a mountain.

5 0

2 years ago