Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º
Answer:
T = 6.0 N
Explanation:
given,
mass of the cord = 0.46 Kg
length of the supports = 7.2 m
time taken to travel = 0.74 s
tension in the chord = ?
using formula for tension calculation
v = 9.73 m/s
now, calculation of tension
T = 6.0 N
The tension in the cord is equal to 6.0 N.
Answer:
10s
Explanation:
If it took Beatrice 25 seconds to complete the race
Distance = 100 meter
Beatrice speed = 100/25
= 4m/s
If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).
Her speed where constant
= 100/20
= 5 m/s
It would take Alice
= 50/5
= 10s
It would take Alice 10s to run $50$ meters.
Answer:
- I have fond the answer
Explanation:
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