What areas of the work force can science be applied too select all that apply

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

8 0

3 years ago

3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second i

Darya [45]

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

(c) Centripetal acceleration = 1849.3031 g

Explanation:

We have given diameter d = 2.30 m

So radius r = $\frac{d}{2}=\frac{2.30}{2}=1.15m$

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec$

(b) Linear speed is given by $v=\omega r=125.6\times 1.15=144.44m/sec$

(c) Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{144.44^2}{1.15}=18141.664m/sec^2$

We know that $g=9.81m/sec^2$

So $18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g$

6 0

2 years ago

When an unstable radioactive atom degrades, what happens to it?

malfutka [58]

it depends on the atom . but in some cases it may explode

7 0

2 years ago

1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line

poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

$v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s$

Hence, his average velocity is 10 m/s.

6 0

2 years ago

Look at the diagram below. What is the size of the overall force on the boat?

inysia [295]

Can i get a picture pls

6 0

2 years ago