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Svetach [21]
3 years ago
7

What increases your ability to see at night

Physics
1 answer:
PtichkaEL [24]3 years ago
8 0
The answer would be increased number <span />of rod cells.Hope this helps :)


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C is the correct answer

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Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa
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1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.
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Read this /https://www.carbonbrief.org/polar-bears-and-climate-change-what-does-the-science-say
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5 0
2 years ago
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee
baherus [9]

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095  and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

4 0
3 years ago
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