Ask question

Ask question

All categories

3 years ago

6

A space probe is orbiting a planet on a circular orbit of radius R and a speed v. The acceleration of the probe is a. Suppose ro

ckets on the probe are fired causing the probe to move to another circular orbit of radius 0.5R and speed 2v. What is the magnitude of the probe’s acceleration in the new orbit?

1 answer:

Mamont248 [21]3 years ago

7 0

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

$a_{new} = 8a.$

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

$a = \frac{v^{2}}{R},$

$a_{new} =\frac{(2v)^{2}}{0.5R},$

$= \frac{4v^{2}}{0.5R}$

$= \frac{8 v^{2}}{R}$

$a_{new} = 8a.$

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

You might be interested in

An object is sitting on the floor. A 22.4 N force is pulling the object to the right and an 11 N force is pulling the object to

Nimfa-mama [501]

answer = 33.4 net force.

5 0

2 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

3 years ago

Which waves can be seen by people?

e-lub [12.9K]

Answer:

Visible waves can be seen by people.

Explanation:

Electromagnetic waves comprises of different kind of waves. It consists of cosmic waves, x-rays waves, gamma waves, ultra violet waves, visible waves, infrared waves, microwaves. Among these waves, depending upon the wavelength of each kind of waves, it is visible and invisible to naked eyes. Thus, visible waves are the one which can be seen by people in naked eyes. As the wavelength of these waves are in comparable to the eye sight, so it can be seen by people.

4 0

2 years ago

A metal begins to emit electrons, as measured in an apparatus similar to the one Hertz used, when exposed to light at a waveleng

balandron [24]

Answer : The work function of this metal is, $5.81\times 10^{-19}J$

Explanation : Given,

Wavelength of light = $342nm=342\times 10^{-9}m$

Formula used :

$E=h\nu_o=\frac{hc}{\lambda}$

where,

E = work function of metal

h = Planck's constant = $6.626\times 10^{-34}Js$

$\nu_o$ = threshold frequency

$\lambda$ = wavelength of light

c = speed of light = $3\times 10^8m/s$

Now put all the given values in this formula, we get the value of work function of this metal.

$E=\frac{hc}{\lambda}$

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{(342\times 10^{-9}m)}$

$E=5.81\times 10^{-19}J$

Therefore, the work function of this metal is, $5.81\times 10^{-19}J$

7 0

3 years ago

1. What are the approximate calorie needs of a

Nimfa-mama [501]

Answer:

D. 2,400 calories

7 0

3 years ago