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Sloan [31]
3 years ago
13

From Earth to the center of our galaxy is about 300,000 light years, meaning that light coming from a star in the center of our

galaxy takes 300,000 years to get to earth. Answer True False
Physics
2 answers:
natima [27]3 years ago
7 0
<span>it takes about about 37,200 years for light to travel 1 light year. So the answer would have to be false. It would take way longer than 300k years

</span>
Alex_Xolod [135]3 years ago
7 0

Answer:

True

Explanation:

1 light year is the distance travelled by light in one year

Distance of 1 light year

1 light year = 3×10⁸×365×24×60×60 m

Speed of light = 3×10⁸ m/s

300,000 light years = 300,000×3×10⁸×365×24×60×60

\text{time}=\frac{\text{Distance}}{\text{Speed}}\\\Rightarrow \text{time}=\frac{300,000\times 3\times 10^8 \times 365\times 24\times 60\times 60}{3\times 10^8}\\\Rightarrow \text{time}=300,000\times 365\times 24\times 60\times 60\ s\\\Rightarrow \text{time}=300,000\ years

So, time taken by light to cover 300,000 light years is 300,000 years

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A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
2 years ago
Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A
Finger [1]

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

r=(100\cos2\theta)\ m

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

v=r\omega

\omega=\dfrac{v}{r}

Where, v = velocity

r = radius

Put the value into the formula

\omega=\dfrac{29.0576}{100\cos30}

\omega=0.335\ rad/s

Hence, The rate at which radar must rotate is 0.335 rad/s.

3 0
3 years ago
explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage​
Talja [164]

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

5 0
2 years ago
find the speed of a rolling ball that travels a distance of 10 m over the top of a smooth table in 2.0 seconds
umka21 [38]
The speed is between 5-15
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2 years ago
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