Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity
, we can use the equation

Solving for
,



b) We can solve for the horizontal component of the velocity
as

or

Answer:
D) D =
, E) (C, D) = (
Explanation:
Part D) two expressions are indicated
3C + 4D = 5
2C +5 D = 2
let's simplify each expression
3C + 4D = 5
4D = 5 - 3C
we divide by 4
D =
The other expression
2C +5 D = 2
2C = 2 - 5D
C =
we can see that the correct result is 1
Part E.
It is asked to solve the problem by the substitution method, we already have
D =
we substitute in the other equation
2C +5 D = 2
2C +5 (5/4 - ¾ C) = 2
we solve
C (2 - 15/4) + 25/4 = 2
-7 / 4 C = 2 - 25/4
-7 / 4 C = -17/4
7C = 17
C =
now we calculate D
D =
D = 5/4 - 51/28
D =
D = - 16/28
D =
the result is (C, D) = (
)
Answer:
E= 2.158× 10*5N/C
Explanation:
K= 8.99×10*9, q= 6×10*-6C, d= 0.5m
E= kq/d*2
E= (8.99×10*9× 6×10*-6)/0.5*2
E= 215760
E= 2.158 ×10*5N/C
In the push-up, the athlete's weight (or force) of 250 N moves through a distance of 20 cm.
By definition,
Work = force * distance.
Therefore, work done in one push-up is

Note that 1 J = 1 N-m.
Answer: 1 J
Your answer would be Option A. Energy is transferred.
I just took this quiz, and can 100% confirm this is the answer.
Sorry this is late, but hopefully this can help future people who visit this question!
Hope this helps! :)