Ask question

Ask question

All categories

2 years ago

9

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

would the solar cell produce if the spacecraft were in orbit around Saturn, 9.5 times as far from the sun?

1 answer:

Mandarinka [93]2 years ago

3 0

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

I_s = I_e * ( r_e / r_s ) ^2

I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

I_s = 11 W / e*a

- We know that P = I*a, hence we have:

P_s = I_s*a

P_s = 11 W / e

Hence, e*P_s = 11 W

You might be interested in

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

tensa zangetsu [6.8K]

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

So

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

5 0

2 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0

2 years ago

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

Drupady [299]

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

6 0

2 years ago

a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a

Eddi Din [679]

Answer:

$a=-3\ m/s^2$

Explanation:

<u>Second Newton's Law</u>

It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by

$F_n=m.a$

The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is

$F_n=30\ N-60\ N=-30\ N$

Thus, the acceleration can be computed by

$\displaystyle a=\frac{F_n}{m}=\frac{-30}{10}=-3\ m/s^2$

$\boxed{\displaystyle a=-3\ m/s^2}$

The negative sign indicates the block is accelerated to the left

7 0

2 years ago

Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th

Stels [109]

This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide. Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after. Subtract this from the the system total and you will have the momentum of the other object after the collision. Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.

Be careful with the velocities. They are vectors, so direction matters. Typically moving to the right is positive (+) and moving to the left is negative (-). It is not clear from your question which direction the objects are moving before and after the collision.

6 0

2 years ago

Read 2 more answers