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Dafna11 [192]
3 years ago
9

A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to

move this charge 50 cm along a path making an angle of 40
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E =  3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶  = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J

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A 5.0 kg mass is suspended from a spring. Pulling the mass down by an additional 10 cm takes a force of 20 N. If the mass is the
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Answer:

0.99 seconds

Explanation:

The problem depicts a simple harmonic motion.

Now, from Hooke's law,

The spring constant is given as;

k =F/x

where:

x is the displacement of the spring's end from its equilibrium position

F is the restoring force exerted by the spring on that end

From the question, F = 20N while x = 10cm = 0.1m. Thus,

K = 20/0.1 = 200 N/m

Now, time to take for the mass to return to its starting point is a period

The period oscillation of the mass is given as;

T = 2π√m/k

Where m = mass = 5kg

T = 2π√(5/200)

T = 0.99 s

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3 years ago
In your own words, describe the purpose of using a chemical equation.
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Chemical equations show how chemicals interact when a reaction occurs. They use symbols and formulas to show the chemical reaction so that chemical names do not need to be written out. They also help keep track of all elements and the number of atoms of each element on each side of the equation.
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Who speaks the line "Lord, what fools these mortals be"?
ivanzaharov [21]
The answer is D.Puck.

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3 0
3 years ago
WOULUJUTUL RECIPECUIUS.
3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

Thus, we get finally,

                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

As we know, nano denoted by letter 'n' equals to 10^{-9}

So the force acting between two objects is 19.73 nN.

7 0
3 years ago
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