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Vitek1552 [10]
3 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]3 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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Answer:

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Explanation:

3 0
2 years ago
. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
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Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

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s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

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j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

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Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

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c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

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3 years ago
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Read more about Light

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As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
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