Answer:
62.6 m/s down
Explanation:
Taking downward to be positive:
Given:
v₀ = 0 m/s
Δy = 200 m
a = 9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (200 m)
v = 62.6 m/s
Answer:
W = 1750 J
Explanation:
It is given that,
Weight of a computer, F = 50 N
A student wants to keep it on the 5th floor which is 35 meters up.
We need to find the work done by her to drag her computer up the stairs. Let it is W. Using the formula of work as follows :
W = F d
Put all the values,
W = 50 N × 35 m
W = 1750 J
So, 1750 J of work is done by her to drag the computer up the stairs.
You don't have to set it up. It's already set up. You only have to identify the correct set-up from the four choices.
The correct one is 'B', and it's a formula that you really should memorize.
You'll be using it in any science class that you take.
Weight of an object = (the object's mass) x (the local acceleration of gravity) .
W = m x g
The object's mass never changes. But its weight does, whenever
it goes to places where the acceleration of gravity is different.
That's why the same astronaut has different weights on Earth
and on the moon.
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Also, by the way, that bit about the "pendulum bob" in this question
may be there mainly to distract and confuse you. Everything I said
up there above the line is true for ANY object ... a pendulum bob,
a gold fish, a newborn baby, a turtle, a tree, a school-bus, a rock,
a house, a jug of milk, a book, a computer, a burrito, a puppy,
a snowball, an airplane, a cigar, or anything else.
Weight anyplace = (mass) x (gravity in that place)
The procedure that makes stars sparkle is the same one that we have outfit for utilization as a weapon: atomic combination.
Stars begin as gigantic districts of gas, for the most part hydrogen. This gas will begin to contract, and it warms up. As it gets, its temperature rises. At the point when the warmth gets sufficiently high, it causes the individual hydrogen iotas to impact and consolidate into helium with the arrival of vitality. This is called fusion.
Answer:
d = 4217 m
Explanation:
Case 1.
Initial velocity, u = 0
Acceleration, a = 2.2 m/s²
Let d₁ is the distance in case 1. Using second equation of motion as follows :
Let d₂ in the distance in case 2. Using second equation of motion as follows :
Total distance,
D = d₁ + d₂
D = 110 + 4107
D = 4217 m
Hence, the total distance covered is 4217 m.