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olga2289 [7]
4 years ago
12

Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

al force on the particle with charge q is

Physics
1 answer:
Gelneren [198K]4 years ago
5 0

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

N_{x} = N cosine 60° + (-N cosine 60°) = 0

for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero (N_{x} = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

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Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25
Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: d_{1x} = 5.2 cos 25^{\circ}=4.7 mi

North: d_{1y} = 5.2 sin 25^{\circ}=2.2 mi

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: d_{2x}=0

North: d_{2y} = 5.0 mi

So the components of the total displacement are

East: d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi

North: d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi

5 0
4 years ago
A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring
egoroff_w [7]

Answer:

4. both blocks will both have the same amount of kinetic energy.

Explanation:

When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.

Spring force is given as:

F=k.\Delta x

where: \Delta x= length of compression in the spring

<u>We know kinetic energy is given by:</u>

KE=\frac{1}{2} m.v^2

Hence the kinetic energy of both the blocks is equal when they are released to move free.

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4 years ago
Why isn't the earth the same distance from sun all year long?
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Answer:

Earth's orbit is elliptical

Explanation:

The earth's orbit is not a straight circle and the sun is not in the very center of it. The orbit is more of a wonky oval with the sun closer to one side of it than the other causing the earth's distance from the sun to vary throughout the year.

7 0
3 years ago
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An 8.7 g bullet is accelerated in a rifle barrel 105 cm long to a speed of 465 m/s. Use the work-energy theorem to find the aver
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Answer:896 N

Explanation:

Given

mass of bullet m=8.7 gm

Length of barrel d=105 cm=1.05 m

final velocity v=465 m/s

initial velocity u=0 m/s

Work done by all the forces is equal to change in kinetic Energy

F\cdot d=\frac{1}{2}\times mv^2-0

F_{avg}\times 1.05=\frac{1}{2}\times 8.7\times 10^{-3}\times 465^2

F_{avg}=\frac{0.5\times 8.7\times 10^{-3}\times 465^2}{1.05}

F_{avg}=895.789 Na\pprox 896 N

6 0
4 years ago
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