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gayaneshka [121]
2 years ago
14

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

ght travels in one year.
Physics
1 answer:
lapo4ka [179]2 years ago
5 0

Answer:

The distance traveled in 1 year is: 3.143*10^{16}ft

Explanation:

Given

s = 982,080,000 ft/s --- speed

t = 32,000,000 s --- time

Required

The distance traveled

This is calculated as:

Speed = \frac{Distance}{Time}

So, we have:

Distance = Speed * Time

This gives:

Distance = 982,080,000 ft/s * 32,000,000 s

Distance = 982,080,000  * 32,000,000ft

Distance = 3.143*10^{16}ft -- approximated

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A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

5 0
3 years ago
Explain why nuclear fission and nuclear fusion release large amounts of energy
levacccp [35]

Answer:

Because of the formula E=mc^2

Explanation:

In this problem we are describing two different processes:

  • Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei
  • Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

E=\Delta mc^2

where:

E is the energy released in the reaction

\Delta m is the mass defect, the difference between the final total mass and the initial total mass

c=3.0 \cdot 10^8 m/s is the speed of light

From the formula, we see that the factor c^2 is a very large number, therefore even if the mass defect \Delta m is very small, nuclear fusion and nuclear fission release huge amounts of energy.

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3 years ago
A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0
OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

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3 years ago
What is work - energy theorem ??​
Elden [556K]

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done.

This idea is expressed in the following equation:

is the total work done

is the change in kinetic energy

is the final kinetic energy

is the initial kinetic energy

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3 0
2 years ago
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A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0
3 years ago
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