Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is
where 
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is
where 
and 
.
Eliminating 
, we have
Solve for .
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
Answer:
85.8 m/s
Explanation:
We know that the length of the circular path, L the plane travels is
L = rθ where r = radius of path and θ = angle covered
Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt
where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path
Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,
θ = θ₀ + ωt = 0 + ωt = ωt
So, v = rdθ/dt + θdr/dt
v = rω + ωtdr/dt
v = (r + tdr/dt)ω
v = (r + tdr/dt)v'/r
v = v' + tv'/r(dr/dt)
v = v'[1 + t(dr/dt)/r]
Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)
So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]
v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]
v = 110 m/s[1 + 11.0 s(-0.02/s)]
v = 110 m/s[1 - 0.22]
v = 110 m/s(0.78)
v = 85.8 m/s
Answer:
but I can tomorrow if you have time can you come to the meeting tonight but yyyy the person who is this and what
Explanation:
gyyyyyyyyyyyyyyyyy the number of the person who is this and what is the
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
