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Tamiku [17]
3 years ago
15

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and directio

n of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components

Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°

Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.

Their resultant is gotten below

Magnitude of resultant = 3.25 - 1.50 = 1.75km

Direction of resultant = north (direction of the bigger vector)

2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.

R = √ vx² + vy²

R = √ 2.20² + 1.75²

R = √ 7.9025

R= 2.1811km.

θ = tan^-1 (vy/vx)

θ = tan ^-1 (1.25/2.20)

θ = tan ^-1 (0.5618)

θ = 29.6°.

The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ

Thus the direction of the vector is 180 + 29.60 = 209.6°

From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.

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A. 5/7h

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C. See attachment below

Explanation:

Find the attachment below for better understanding.

7 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

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3 0
3 years ago
A coil is wrapped with 332 turns of wire on the perimeter of a circular frame (of radius 30 cm). Each turn has the same area, eq
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Answer:

E=84.5V

Explanation:

From the question we are told that:

Number of Turns N=332turns

Radius r= 30cm

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Time t=63ms

Generally the equation for Magnetic Field is mathematically given by

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Generally the Flux at 332 turns is  mathematically given by

 \phi=N*A*B

Generally the equation for Area of coil is mathematically given by

 A=\pi*r^2

 A=\pi*(r*10^{-2})^2

Since

 \phi=332*\pi*(\theta*10^{-2})^2*B

Therefore

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Generally the equation for emf Magnitude is mathematically given by

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4 0
2 years ago
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Answer:

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2 years ago
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A 2-kg object is moving horizontally across a frictionless surface with a speed of 4 m/s. How much force is required to keep the
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Answer:

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No force is necessary for the object to keep its speed and direction on a frictionless surface.

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