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AlladinOne [14]

2 years ago

15

an object that is 8 cm tall is located 2.5 m in front of a plane mirror. The image formed by the mirror appears to be. Find di a

nd whether behind or in front of mirror

1 answer:

UkoKoshka [18]2 years ago

6 0

The image is at 2.5 m behind the mirror (virtual image)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

where

f is the focal length of the mirror

$d_o$ is the distance between the object and the mirror

$d_i$ is the distance between the image and the mirror

For a plane mirror, the focal length is infinite:

$f=\infty$

And in this problem, the distance of the object from the mirror is

$d_o = 2.5 m$

Substituting and solving for $d_i$ , we find the location of the image:

$0=\frac{1}{d_o}+\frac{1}{d_i}\\d_i = -d_o = -2.5 m$

And the negative sign means that the image is located behind the mirror, so it is virtual.

Learn more about reflection and refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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Answer:

se me hace que son 8 protones

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3 years ago

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A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

Marizza181 [45]

Answer:

1.7×10^5 ms-1

Explanation:

From

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v= E/B= 110×10^3/0.6

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The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box

Anna [14]

We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

$F_f=\mu N$

Where.

$\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}$

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

$\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}$

Adding the forces in the y-direction we get:

$\Sigma F_y=N-mg_y$

Since there is no movement in the y-direction the sum of forces must be equal to zero:

$N-mg_y=0$

Now we solve for the normal force:

$N=mg_y$

To determine the y-component of the weight we will use the trigonometric function cosine:

$\cos 40=\frac{mg_y}{mg}$

Now we multiply both sides by "mg":

$mg\cos 40=mg_y$

Now we substitute this value in the expression for the normal force:

$N=mg\cos 40$

Now we substitute this in the expression for the friction force:

$F_f=\mu mg\cos 40$

Now we substitute the given values:

$F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40$

Solving the operations:

$F_f=15.01N$

Therefore, the force of friction is 15.01 Newtons.

3 0

10 months ago

Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours

MariettaO [177]

With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration.

5 0

2 years ago

Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti

Triss [41]

Answer:

<em>The total potential (magnitude only) is 11045.45 V</em>

Explanation:

<u>Electric Potential </u>

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

$\displaystyle V=\frac{kq}{r}$

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

$\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V$

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

$V_2=-22090.91\ V$

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

$V_3=+22090.91\ V$

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

$\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V$

The total electric potential in A is

$V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V$

$V=-11045.45 \ V$

The total potential (magnitude only) is 11045.45 V

8 0

2 years ago