A 10.0 cm3 sample of copper has a mass of 89.6

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

Mandarinka [93]

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

3 0

2 years ago

The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0

2 years ago

Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

$V_2 = \frac{V_1}{2}$

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

8 0

2 years ago

A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. you decide to

Zinaida [17]

Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision

hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)

7 0

2 years ago