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2 years ago

A convex mirror is placed to the right of an object. The image formed by the mirror will be a

Physics

2 answers:

Harrizon [31]2 years ago

6 0

I think it’s C but I’m not too sure

Amanda [17]2 years ago

4 0

<h3><u>Answer;</u></h3>

C.) Virtual Image that appears on the right side of the mirror.

A convex mirror is placed to the right of an object. The image formed by the mirror will be a <u><em>Virtual Image that appears on the right side of the mirror. </em></u>

<h3><u> Explanation;</u></h3>

<em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.

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a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

shutvik [7]

Isothermal Work = PVln(v₂/v₁)

PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = $e^{0.3098}$

v₂ = 19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L Option b.

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2 years ago

A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni

dimaraw [331]

Answer:

The magnitude of the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})$

Put the value into the formula

$E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})$

$E=5.75\ N/C$

The direction is toward positive x- axis.

Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.

7 0

2 years ago

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

$W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m$

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

$W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J$

So, the new work is more than 130 J.

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2 years ago

Electron configuration is responsible for the make up of the chemical properties of elements

allochka39001 [22]

This is True

I hope i helped

3 0

2 years ago

Read 2 more answers

A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs

Contact [7]

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

4 0

3 years ago