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Montano1993 [528]
2 years ago
10

vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone

nts of the vector?
Physics
1 answer:
Reika [66]2 years ago
5 0

We have vector (17.9*cos80^{0},17.9*sin80^{0})

Therefore,

x component = 17.9 * cos80 degree = 3.108

y component = 17.9 * sin80 degrees = 17.628

<h3>What is a vector?</h3>

An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

If the magnitude and direction of two vectors match, they are the same vector. This shows that if we move a vector to a different location without rotating it, the final vector will be the same as the initial vector. The vectors that denote force and velocity are two examples. The direction of force and velocity are both fixed. The size of the vector would represent the force's strength or the velocity's corresponding speed.

To know more about vectors, visit:

brainly.com/question/12937011

#SPJ4

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Gnom [1K]

|Momentum| = (mass) x (speed)

225 kg-m/s =(50kg) x (speed)

Divide each side by (50kg):  Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .

Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.

8 0
3 years ago
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A 5 kg rock is raised 28 m above the ground level. What is the change in its potential energy?
marin [14]
Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
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Which of the following best describes the use of a renewable resource? (2 points) Most power plants burn fossil fuels to generat
kolbaska11 [484]

Which of the following best describes the use of a renewable resource?

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There are areas in California that use the heat from deep in the Earth to generate electricity.

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5 0
3 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

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4 0
4 years ago
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Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

            F = m a

the electric force is

           F = q E

we substitute

          q E = m a

           a = qE / m

the mass of the particle is m = 2.00 10-16 kg

           a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

           a = 1,616 m / s²

           

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

            t = 2v_{oy} / a_y

let's calculate

           t = 2 0.6139 10⁵ / 1.616

           t = 7.597 10⁴s

in this time the particle travels a horizontal distance

           x = v₀ₓ t

           x = 0.8145 10⁵ 7.597 10⁴

           x = 6.19 10⁹ m

the particle falls off the plate

4 0
3 years ago
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