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3 years ago

When triggers nuclear fusion in stars

1 answer:

5 0

Fusion in stars are usually the result of gravity. Once a mass of hydrogen accumulates enough mass, the gravity of all that mass compresses the core of the star to the point that the hydrogen atoms there begin fusing into helium. The <span>process then cascades outward, and the end result is a star. Hopes ths Helps. Also mark me brainliest since I was first

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The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

4 years ago

Read 2 more answers

Determine the number of ways to throw two die and get the number 11, as well as the probability of getting 11.

Ivenika [448]

Answer:

1/18

Explanation:

The number of ways in which two die can be thrown is the sample space of the experiment

(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)

(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)

(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)

(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)

(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)

(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)

From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)

$\text{Probability of getting 11}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\\\Rightarrow \text{Probability of getting 11}=\frac{2}{36}\\\Rightarrow \text{Probability of getting 11}=\frac{1}{18}=0.056$

∴ Probability of getting 11 is<u> 1/18</u>

8 0

4 years ago

What do lenses do?

Dovator [93]

A convex lens makes light rays converge (come together) at the focal point or focus. The distance from the center of the lens to the focal point is the focal length of the lens. Convex lenses are used in things like telescopes and binoculars to bring distant light rays to a focus in your eyes.

3 0

4 years ago

Read 2 more answers

earth orbits the sun once every 365.25 days. Find the average angular. speed of Earth about the sun.Answer in units of rad/s.

anzhelika [568]

Answer:

$1.99\cdot 10^{-7} rad/s$

Explanation:

The average angular speed of the Earth about the Sun is given by:

$\omega = \frac{2 \pi}{T}$

where

$2 \pi$ rad is the total angle corresponding to one revolution of the Earth around the Sun

T is the orbital period of the Earth

The orbital period of the Earth is 365.25 d. We must convert it into seconds first:

$T=365.25 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=3.16\cdot 10^7 s$

And by substituting into the equation above, we find the average angular speed:

$\omega=\frac{2 \pi rad}{3.16\cdot 10^7 m/s}=1.99\cdot 10^{-7} rad/s$

7 0

3 years ago

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You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

$h=r-R=20395.3km-3389.5 km=17005.8 km$

6 0

3 years ago