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Viktor [21]
3 years ago
11

You are observing a star that is 250 million light years away. From the starlight you are observing now, the appears to be 30 mi

llion years old and you determine that the star has a lifetime of 120 million years. How long will it be until Earth receives light from the supernova that occurs at the end of the life of the star
Physics
1 answer:
Romashka [77]3 years ago
4 0

What ever games your playing it wont work you cant defeat me hahhahha oh I know :O but he can AHHHH :D D: NOOOO IMPOSSIBLE.

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You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use
Yuri [45]
P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]
3 0
3 years ago
How does the distance traveled by the coin compare to its displacement after ten flips?
Mkey [24]

Answer:

To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.

Explanation:

hope helps

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3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
Ray Of Light [21]
There has to be 6 N for a reactant


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3 years ago
You hear an _____ when sound bounces off a hard surface.
Aleksandr [31]
I think b but I’m not completely sure
7 0
4 years ago
Read 2 more answers
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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