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olga nikolaevna [1]
3 years ago
6

A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

4 cm. Calculate the focal length of the mirror.
Physics
1 answer:
Aloiza [94]3 years ago
4 0

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

f=\dfrac{R}{2}

f=\dfrac{-31}{2}

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
A submarine travels 25 km/h north for 3.2 hours. What is its displacement?
Vika [28.1K]

displ = velocity x time

25 x 3.2 = 75+5 km north.

7 0
3 years ago
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Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t
umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2>F = 1991.05 N</h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

mg \times L = F \times d

so we have

18.14 \times 9.8 \times 0.56 = F \times (0.05)

F = 1991.05 N

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3 years ago
In a given chemical reaction the energy of the products is less than the energy of the reactants. Which statement is true for th
Westkost [7]

Answer:

B. Energy is released in the reaction.

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3 years ago
On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

                                 T  = 2*pi*sqrt(k / m )    ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

                                 F_net = k*x - m*g_x = 0

                                 (k / m) = (g_x / x)    .... 2

- substitute Eq 2 into Eq 1:

                                  2*pi / T = sqrt ( g_x / x )

                                   g_x = (2*pi / T )^2 * x

- Evaluate g_x:

                                  g_x = (2*pi / (19 / 11) )^2 * 0.226

                                  g_x = 3.0 m / s^2

                                 

                       

3 0
3 years ago
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