Answer:
F.
Explanation:
Here in the question the mass of the pulley is zero, hence, the tension in the cable throughout is same.
magnitude of tension in rope 1 is
T1= F
Hence the tension T1 is rope 1 is F.
The difference in emf between the primary coil of a transformer and the secondary coil is directly related to the relative numbe
1 answer:
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Answer:
V = 90.51 m/s
Explanation:
From the given information:
Initial speed (u) = 0
Distance (S) = 391 m
Acceleration (a) = 18.9 m/s²
Using the relation for the equation of motion:
v² - u² = 2as
v² - 0² = 2as
v² = 2as
v = 121.57 m/s
After the parachute opens:
The initial velocity = 121.57 m/ss
Distance S' = 332 m
Acceleration = -9.92 m/s²
How fast is the racer can be determined by using the relation:
V = 90.51 m/s
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²
Explanation:
The expression left corresponds to an oscillatory movement (MAS), the speed is defined by
v = dr / dt
the function of position
r = 2 cos πt i^ + 3 sin πt j^
let us note that it is a movement in two dimensions
let's perform the derivative
v = -2π sin πt i^ + 3π cos πt j^
we evaluate this expression for t = 0.25 s, remember that the angle is in radians
v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^
v = (-4.44 i^ + 6.66 j^ ) m/s
To calculate the mean acceleration we use the expression
a = () / Δt
indicates that the time is the first 3 s
we look for the initial velocity t = 0 s
v₀ = 0 i ^ + 3π j ^
we look for the fine velocity, t = 3 s
v_f = - 2π sin (π 3) + 3π cos (π 3) j ^
v_f = 0 i ^ - 3π j ^
we calculate the average acceleration
Δt = (3 -0) = 3 s
a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3
a_average = (0 i ^ -2π j ^ ) m/s²