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Citrus2011 [14]
3 years ago
15

Look at the vector below. It is at a 58 degree angle and has a magnitude of 35.

Physics
1 answer:
Alexxandr [17]3 years ago
3 0
  • x-component: 18.55;
  • y-component: 29.68.
<h3>Explanation</h3>

Consider the vector as the hypotenuse of a right triangle. Refer to the diagram attached. The x and y components are the two legs.

The x-component is adjacent to the 58° angle.

x\text{-component} = \text{Hypotenuse} \cdot \cos{\text{Angle} = 35 \times \cos(58\textdegree{}}) = 18.55.

The y-component is opposite to the 58° angle.

y\text{-component} = \text{Hypotenuse} \cdot \sin{\text{Angle} = 35 \times \sin(58\textdegree{}}) = 29.68.

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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
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To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

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\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

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Recall that the speed of light is equivalent to

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Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

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B = \frac{r}{2C^2} \frac{dE}{dt}

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A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio
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We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

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