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2 years ago

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3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in

the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.

1 answer:

Eduardwww [97]2 years ago

4 0

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

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A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.

taurus [48]

In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

4.9 N-m = (force) x (0.89 m)

Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)

= 5.506 N .

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2 years ago

Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b

Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

$\Delta ACB$

$tan 30 = \frac{8.3}{h}$

$h= \frac{8.3}{tan 30} = 14.376 m$

therefore

height of keisha is $= h + 1.5 m$

= 14.376 + 1.5

$= 15.876 \simeq 16 m$

therefore option c is correct

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2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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2 years ago

Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len

love history [14]

Answer:

(c) +6.67

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

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Distance from second lens is 10 cm to the right

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm$

The final image will appear as +6.67 cm

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2 years ago

Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

Tema [17]

Note: The diagram referred to in the question is attached here as a file.

Answer:

The magnitude of the magnetic field is $B = \frac{0.071 \mu I}{d}$

Explanation:

The magnetic field can be determined by the relationship:

$B = \frac{\mu I}{2\pi R}$ ...............(1)

Were I is the current flowing through the wires

The distance R from point 1 to m is calculated using the pythagora's theorem

$R = \sqrt{d^{2} + (2d)^{2} }$

$R = \sqrt{5d^{2} } \\R = d\sqrt{5}$

Substituting R into equation (1)

$B = \frac{\mu I}{2\pi d\sqrt{5} }$

$B = \frac{0.071 \mu I}{d}$

3 0

2 years ago