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dezoksy [38]
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.
Physics
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}

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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
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