Due in a few minutes plzzzz help! Will mark brainliest
1 answer:
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Answer:
Explanation:
Given that, .
Mass of ladder is 51kg
Then, it weight is
WL = mg = 51 × 9.81 = 500.31N
This weight will act at the midpoint of the ladder
Length of ladder is 15m
The ladder makes an angle 55°C with the horizontal
An object whose mass is 81kg is at 4m from the bottom of the ladder
Then, weight of object
Wo = mg = 81 × 9.81 = 794.61 N
Using newton second law
Check attachment
Ng is normal force on the ground
Ff is the horizontal frictional force
Nw Is the normal force on the wall
ΣFy = 0
Ng = Wo + WL
Ng = 794.61 + 500.31
Ng = 1294.92 N
Also
ΣFx = 0
Ff — Nw = 0
Then,
Ff = Nw
Now taking moment about point A.
Check attachment
using the principle of equilibrium
Sum of clockwise moment equals to sum of anti-clockwise moment
Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object
Clockwise = Anticlockwise
Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15
794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15
1589.22 + 1876.163 = 12.99•Nw
3465.383 = 12.99•Nw
Nw = 3465.383 / 12.99
Nw = 266.77 N
Since, Nw = Ff
Then, Ff = 266.77N
the horizontal force exerted by the ground on the ladder is 266.77 N
Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J