Ask question

Ask question

All categories

4 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]4 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

___________ relates the quantities of pressure, volume, temperature, and quantitiy of moles.

Gnoma [55]

C.) The relationship between volume, temperature, pressure and quantity of moles is given by ideal gas law. Mathematically given by PV = nRT.

5 0

3 years ago

An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Nonrel

lozanna [386]

Answer:

<em>1.187 x 10^8 m/s</em>

<em></em>

Explanation:

the potential of the electric field V = 40 kV = 40000 V

the charge on an electron e = 1.6 x 10^-19 C

The energy of an accelerated electron in an electric field is given as

E = eV

E = 1.6 x 10^-19 x 40000 = 6.4 x 10^-15 J

This energy is equal to the kinetic energy with which the electron moves, according to the conservation of energy.

The kinetic energy = $\frac{1}{2}mv^{2}$

where

m is the mass of the electron = 9.109 x 10^-31

v is the speed of the electron.

Equating the energy, we have

6.4 x 10^-15 = $\frac{1}{2}*9.109*10^-31*v^{2}$

6.4 x 10^-15 = 4.55 x 10^-31 $v^{2}$

$v^{2}$ = 1.41 x 10^16

$x^{2} v = \sqrt{1.41*10^{16}}$ = <em>1.187 x 10^8 m/s</em>

3 0

3 years ago

Al colocar un objeto delante de una lente convergente se genera una imagen real invertida y de doble tamaño, sabiendo que dicha

alexira [117]

Answer:

30 cm

Explanation:

To solve this problem, we use the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we know that

q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image

We also know that the size of the image is twice that of the object, so the magnification is 2:

$M=-\frac{q}{p}=2$

where M is the magnification of the lens. Solving this equation for p,

$p=-\frac{q}{2}=-\frac{-30}{2}=15 cm$

So, the distance of the object from the lens is 15 cm.

Now we can finally solve the lens equation to find f, the focal length:

$\frac{1}{f}=\frac{1}{15}+\frac{1}{-30}=\frac{1}{30}\\\rightarrow f=30 cm$

5 0

4 years ago

Help me out please I need to find this out quick

photoshop1234 [79]

He got u too bad for the late reply to the question I have;)

6 0

3 years ago

Read 2 more answers

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

Scilla [17]

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

Acceleration due to gravity $g=9.8m/sec^2$

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

8 0

4 years ago