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Diano4ka-milaya [45]
2 years ago
14

As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65

degrees above the horizontal 45 N at an angle of 65 degrees above the horizontal.
500 kg
The horizontal component of the force acting on the crate is?
Physics
1 answer:
kaheart [24]2 years ago
4 0

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

\theta = Angle at which the force is being applied = 65^{\circ}

Horizontal component of force is given by

F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}

The horizontal component of the force acting on the crate is 19.01 N.

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3 years ago
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
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(a) The emf of a rotating coil with N turns is given by:

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N: turns

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A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

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(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

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{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}

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