Power is equivalent to the energy released divided by the time taken to release the energy. true false

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a

Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0

2 years ago

What kind of energy does a flying bullet have?

Musya8 [376]

It mainly travels by kinetic energy

3 0

3 years ago

Read 2 more answers

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago

What are the components of the "Earth Radiation Budget?"

Alex777 [14]

Answer:

The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.

Explanation:

I hope this helps also I hope you have a great day and a new year.

4 0

2 years ago

The mass of a certain neutron star is 2.5x10^30kg and the radius 7000m. what is the force of gravity on a 1kg object of the surf

Makovka662 [10]

Answer:

3.42N

Explanation:

*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*

Use the equation: F = (GMm)/(r^2)

F = force of gravity

G = gravitational constant (6.7x10^-11)

M = mass1 (2.5x10^30kg)

m = mass2 (1kg)

r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

F = (1.675x10^20) / (4.9x10^7)

F = 3.4183673x10^12

F = 3.42N

8 0

3 years ago