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3 years ago

Which describes an image that a concave mirror can make? Which describes an image that a concave mirror can make?

Physics

1 answer:

Shkiper50 [21]3 years ago

3 0

Answer:

the image can be rather real or virtual

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Select all that apply. Which of the following would increase the conductivity of a connecting wire?

lord [1]

What are your choices

4 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

3 years ago

Why are mercury and the earths moon so cratered?

Ksenya-84 [330]

They are cratered because of meteoroids, neither of them have an atmosphere to disintegrate the meteors, and neither of them have water to erode the craters.

8 0

3 years ago

First, find the magnitude of v⃗ , that is, the speed v of the two-car unit after the collision.

Rufina [12.5K]

The magnitude of v⃗ is {sqrt(m1v1)2+(m2v2)2/ m1+m2}, that is, the speed v of the two-car unit after the collision.

<h3>What is collision?</h3>

In physics, collisions occur when particles, aggregates of particles, or solids come close to each other, interact and affect each other.
Collisions are of three types Fully elastic collision, inelastic collision and Perfectly inelastic collision.
Multiply the mass of the second object by its velocity.
For example, if the weight is 1,000 and the speed is -30 meters per second, then its momentum is 30,000 kg meters per second.
Add the two velocities together to determine the direction the object will move after a collision.
So the formula for determining the size of a vector (in 2D space) is v = (x, y).|v| = √(x2 + y2).
This formula is derived from the Pythagorean theorem.
The formula V = (x, y, z) that determines the size of a vector (in 3-dimensional space) is:|V| = √(x2 + y2 + z2)

To learn more about collision from the given link :

brainly.com/question/13138178

#SPJ4

3 0

1 year ago

Which one is series and which one is parallel ( i give 20 points who answer )

slega [8]

Answer:

figure (a) L1 is series and L2 is parallel to L3

figure (b) L2 and L3 are parallel to L1

4 0

2 years ago