Ask question

Ask question

All categories

2 years ago

14

A rocket with mass m traveling with speed v0 along the x axis suddenly shoots out one-third its mass parallel to the y axis (as

seen from an observer at rest) with speed 2v0. Find the x, y, and z components of the final velocity vector of the remaining 2/3 of the rocket's mass.

1 answer:

SpyIntel [72]2 years ago

7 0

Answer:

$\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}$

Explanation:

Initially the rocket velocity is only on the x-axis. After the rocket suddenly shoots out one-third its mass parallel to the y axis, its velocity is in the x and y axes. Thus, According to momentum conservation principle, we have:

$mv_0=\frac{2}{3}mv_{x}\\v_x}=\frac{3}{2}v_0\\\\0=\frac{1}{3}m2v_0+\frac{2}{3}mv_{y}\\v_y=-\frac{2}{3}m(v_0)[\frac{3}{2m}]\\v_y=-v_0\\\\(0)_z=(0)_z\\v_z=0$

So, the velocity vector of the remaining rocket's mass is:

$\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}$

You might be interested in

Travels 11,000 feet along a dark desert highway if the car averages 84 mph find the amount of time to cover this distance

laila [671]

The time taken by traveler to cover the distance is,

$t=\frac{d}{v}$

Substitute the known values,

$\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}$

Therefore, the time taken by traveler to cover the distance is 89.3 s.

5 0

10 months ago

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

asambeis [7]

Answer:

$I=0.0987kg.m^2$

Explanation:

From the question we are told that:

Mass $M=1.80kg$

Deviation $d=0.250$

Time $t=0.940s$

Generally the equation for moment of inertia is mathematically given by

$I=\frac{T}{2\pi}^2(mgd)$

$I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)$

$I=0.0987kgm^2$

5 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

2 years ago

HELP ASAP PLS!<br><br> Explain Newton’s 3 laws of motion by using the example of a rollercoaster.

dybincka [34]

For every action there is an equal and opposite reaction." So that applies to a roller coaster, between the ride vehicles and the track. When a ride goes up and down the hill, it creates different forces onto the track.

8 0

3 years ago

calculate the force required to take away a flat corcular plate of radius 0.01m from the surface of water. the surface tention o

leonid [27]

Answer:

$Force = 0.0047175\ N$

Explanation:

Given

$T = 0.075N/m$ --- Surface Tension

$r = 0.01m$ --- Radius

Required

Determine the required force

First, we calculate the circumference (C) of the circular plate

$C= 2\pi r$

$C= 2 * \frac{22}{7} * 0.01m$

$C= \frac{2 * 22 * 0.01}{7}m$

$C= \frac{0.44}{7}m$

$C= 0.0629 m$

The applied force is then calculated using;

$Force = C * T$

$Force = 0.0629m * 0.075N/m$

$Force = 0.0047175\ N$

8 0

2 years ago