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Vitek1552 [10]
4 years ago
5

If v lies in the first quadrant and makes an angle Ï/3 with the positive x-axis and |v| = 4, find v in component form.

Physics
1 answer:
Alexeev081 [22]4 years ago
4 0

Answer:

v=

Explanation:

We are given that

Magnitude of vector v=\mid v\mid =4

v lies in the first quadrant

\theta=\frac{\pi}{3}

v_x=\mid v\mid cos\theta

v_y=\mid v\mid sin\theta

Substitute the values then we get

v_x=4cos\frac{\pi}{3}

v_x=4\times \frac{1}{2}=2

cos\frac{\pi}{3}=\frac{1}{2}

v_y=4\times sin\frac{\pi}{3}=4\times \frac{\sqrt 3}{2}=2\sqrt 3

sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Therefore, the vector v in component form=

v=

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I need help with this question how to solve it for Brass and Cooper
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Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
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Using lens formula

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Hence, The image distance is 20.0 cm.

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