Answer:
Correct answer: F = 50 N weight of backpack
Explanation:
Given:
W = 200 J work against gravity
h = 4 m
F = m g = ?
g = 10 m/s²
The work done is a measure of change in this case of potential energy
W = Δ Ep = m g h = F · h ⇒ F = W / h =
F = W / h = 200 / 4 = 50 N
F = 50 N
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Answer:
Explanation:
We shall represent all the displacement in terms of vector . We shall consider east as i , and north as j . south of west direction will be represented by - i - j .
displacement of 1.9 km due east
D₁ = 1.9 i
vector representing south of west = - i - j
unit vector = - i - j / √ 2
7.2 km south of west = 7.2 ( - i - j ) / √ 2
D₂ = - 5.09 ( i + j )
Total displacement
= D₁ + D₂
= 1.9 i - 5.09 ( I + J )
D = - 3.19 i - 5.09 j .
magnitude of D = √ ( 3.19² + 5.09² )
= 6 km .
Direction of D
Tanθ = 5.09 / 3.19 = 1.59
θ = 58°
So direction will be 58° south of west .
To reach the starting point , he shall have to go in opposite direction .
So he shall have to go in the direction of north of east at angle 58° by a displacement of 6 km .
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
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