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4 years ago

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An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista

nce? Use the equation . A. 16.7 cm B. -50.0 cm C. 50.0 cm D. -16.7 cm

2 answers:

ioda4 years ago

6 0

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

tamaranim1 [39]4 years ago

3 0

Answer:

The other person was right, the answer is 50.0

Explanation:

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Define self-actualization.

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Answer:the realization or fulfillment of one's talents and potentialities, especially considered as a drive or need present in everyone.

Explanation:

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3 years ago

Read 2 more answers

A proton is first accelerated from rest through a potential difference V and then enters a uniform 0.750-T magnetic field orient

galben [10]

The magnetic force acting on a charged particle moving perpendicular to the field is:

$F_{b}$ = qvB

$F_{b}$ is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

$F_{c}$ = mv²/r

$F_{c}$ is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set $F_{b}$ equal to $F_{c}$ and solve for v:

qvB = mv²/r

v = qBr/m

Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:

W = KE

W is work, KE is kinetic energy

W = Vq

KE = 0.5mv²

Therefore:

Vq = 0.5mv²

Substitute v = qBr/m and solve for V:

V = 0.5qB²r²/m

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

B = 0.750T

q = 1.60×10⁻¹⁹C (proton charge)

r = 1.84×10⁻²m

Plug in the values and solve for V:

V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷

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4 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

3 years ago

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m a

balu736 [363]

Answer:

The value is $\lambda = 214.3 \ nm$

Explanation:

From the question we are told that

The slit separation is $d = 3.00 * 10^{-5} m$

The distance of the screen is $D = 2.00\ m$

The order of fringe is n = 7

The path difference is $y = 10.0 \ cm = 0.1 \ m$

Generally the path difference is mathematically represented as

$y = \frac{n * \lambda * D}{ d}$

=> $0.1 = \frac{7 * \lambda * 2.00 }{ 3.00 * 10^{-5}}$

=> $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=> $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=> $\lambda = 2.143 *10^{-7} \ m$

=> $\lambda = 214.3 \ nm$

6 0

3 years ago

Wan and Nurul sat on a see-saw. The see-saw is imbalanced because Wan’s mass is 45 kg while Nurul’s mass is only 30 kg. Suggest

Alex Ar [27]

We have that the see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

From the Question we are told that

Wan’s mass is $M_w=45 kg$

Nurul’s mass is $M_n= 30 kg.$

Generally

The Will be balance when the weight on both sides of the see-saw are equal

$M_w=M_n+x$

$45=30+x$

$x=45-30$

$x=15kg$

In conclusion

The see-saw will be balance if a weight of $x=15kg$ is added to Nural's side of the see-saw

$x=15kg$

For more information on this visit

brainly.com/question/22255610

5 0

3 years ago