a person starts at a position of 2 meters and finishes at a postion of 25 meters. the trip takes 4.5 seconds. what is the person
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Answer: option C. 43.9°
Explanation:
Refer to the diagram. The tension in the two strings is equal T₁=T₂=T = 3 N
The strings support a frame weighing, W= 4.26 N
We will write the tension in the strings as the sum of their horizontal and vertical components.
The horizontal components balance each other and cancel out.
T₁ cos θ = T₂ cos θ
On the other hand, The vertical component would support the weight of the frame.
2 T sin θ = W
2 × 3 N × sin θ = 4.16 N
sin θ = 0.693
⇒ θ = sin ⁻¹(0.693) = 43.9°
Hence, the angle made by two wires with the horizontal is 43.9°.
The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by
so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:
where is the acceleration due to gravity. Then we find that
, at which point we find
.
Refer to the diagram shown below.
m = 8.5 kg, the mass of the cat
F = 41.0 N, the force acting up the incline on the cat
θ = 19°, the inclination of the ramp to the horizontal
u = 1.9 m/s, the initial speed along the ramp of the cat
s = 2 m, the length of the ramp
g = 9.8 m/s²
Friction is negligible.
The force F is the component of the cat's weight along the ramp.
F = mg sinθ
= (8.5 kg)*(9.8 m/s²) sin(19°)
= 27. 1198 N
The net force pushing the cat up the ramp is
41.0 - 27.1198 = 13.88 N
If the acceleration of the cat up the ramp is a, then
(8.5 kg)*(a m/s²) = 13.88 N
a = 1.6329 m/s²
Let v = the velocity at the top of the ramp.
Then
v² = u² + 2as
v² = (1.9 m/s)² + 2*(1.6329 m/s²)*(2 m) = 10.1416 (m/s)²
v = 3.185 m/s
Answer: 3.185 m/s
m = 8.5 kg, the mass of the cat
F = 41.0 N, the force acting up the incline on the cat
θ = 19°, the inclination of the ramp to the horizontal
u = 1.9 m/s, the initial speed along the ramp of the cat
s = 2 m, the length of the ramp
g = 9.8 m/s²
Friction is negligible.
The force F is the component of the cat's weight along the ramp.
F = mg sinθ
= (8.5 kg)*(9.8 m/s²) sin(19°)
= 27. 1198 N
The net force pushing the cat up the ramp is
41.0 - 27.1198 = 13.88 N
If the acceleration of the cat up the ramp is a, then
(8.5 kg)*(a m/s²) = 13.88 N
a = 1.6329 m/s²
Let v = the velocity at the top of the ramp.
Then
v² = u² + 2as
v² = (1.9 m/s)² + 2*(1.6329 m/s²)*(2 m) = 10.1416 (m/s)²
v = 3.185 m/s
Answer: 3.185 m/s