To determine the distance of the light that has traveled given the time it takes to travel that distance, we need a relation that would relate time with distance. In any case, it would be the speed of the motion or specifically the speed of light that is travelling which is given as 3x10^8 meters per second. So, we simply multiply the time to the speed. Before doing so, we need to remember that the units should be homogeneous. We do as follows:
distance = 3x10^8 m/s ( 8.3 min ) ( 60 s / 1 min ) = 1.494x10^11 m
Since we are asked for the distance to be in kilometers, we convert
distance = 1.494x10^11 m ( 1 km / 1000 m) = 149400000 km
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
No grooming your snail while driving. Just ask Spongebob. Oh by the way, what are the options?
complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
