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faust18 [17]
3 years ago
9

a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector

Physics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

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4 years ago
A boy rides a sled down a steep, snow.covered hill. Which of
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a. Potential energy decreases and Kinetic energy increases

Explanation:

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3 years ago
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A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

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We have the final answer as

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Hope this helps you

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3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':
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Answer:

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Considering kinematics formula for final velocity as

v^{2}=u^{2}+2as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

v=\sqrt{u^{2}+2as}

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s

Also, v=u+at and making t the subject of the formula

t=\frac {v-u}{a}

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s

Therefore, it needs 7.6 seconds to travel

7 0
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