The answer would be (A) Protons
Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:


I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
The solution for the problem is:
1 Watt = 1 Joule per second
1 Watt*second = 1 Joule
a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds
15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000
Watt*second
54,000,000 Watt*second = ? Joules
54,000,000 Joules / second = 54,000,000 Watts
Kinetic energy is given by



Above is the kinetic energy of a moving car
Answer:
cheaper production
reduced unemployment
Explanation:
I'm sorry if this might be incorrect
The achievements made by corporations were cheaper production and reduced unemployment.