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3 years ago

What is the magnitude and direction of an electric field that exerts a 2.00×10−5N upward force on a –1.75μC charge?

Physics

1 answer:

Sophie [7]3 years ago

3 0

Answer:

E = - 11.4 N/C, Directed Downward

Explanation:

Force = 2.00×10⁻⁵N , charge Q = –1.75μC = -1.75 ×10⁻⁶ C

To Find: Electric Field E = ? and Direction=?

Sol

we Have

E = F / Q

E = 2.00×10⁻⁵N / -1.75 ×10⁻⁶ C

E = - 11.428 N/C (-ve sign shows that it is Negative Electric Field around a negative charge and will attract the Positive Field)

As given an upward force on negative charge so direction of the Electric Filed will be Downward (as Electric field is opposite to direction of force on negative charge)

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M=-2 the imagine equals what?

WITCHER [35]

Answer:

Image is twice as large as the object, and inverted

Explanation:

When an object is placed in front of a mirror, the mirror produces an image.

The magnification of the image is a number telling how much the size of the image is enlarged/diminished with respect to the object.

It is given by

$M=\frac{y'}{y}$

where

M is the magnification

y' is the size of the image

y is the size of the object

In this problem,

M = -2

This means that:

$y'=My\\y' = -2y$

So, we can conclude the following:

- The size of the image is twice the size of the real object

- The image is also inverted, because of the presence of the negative sign in the equation

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3 years ago

If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

elena-14-01-66 [18.8K]

Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>

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3 years ago

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Cars A and B are racing each other along the same straight road in the following manner:

zysi [14]

Answer:

$\frac{x_{o}}{v_B-V_A} =t$

Explanation:

Represent the car's position as a function

$x_o=$ "head start"

$x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt \\$

$Remember: v_B>v_A$

"cathching up means" that $x_A(t)=x_B(t)$

$x_{o} + v_{A}t =v_Bt\\x_{o} = v_Bt -v_{A}t\\x_{o} = t(v_B-V_A)\\\frac{x_{o}}{v_B-V_A} =t, where \ v_B>v_A$

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3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0

3 years ago

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An object falls from the top of a building that is 25 m high. Air resistance is negligible.

Vlada [557]

The velocity of the object s calculated as 22.1 m/s.

<h3>What is the speed of the object?</h3>

Given that we can write that;

v^2 = u^2 + 2gh

Now u = 0 m/s because the object was dropped from a height

v^2 = 2gh

v = √2 * 9.8 * 25

v = 22.1 m/s

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