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3 years ago

5

A concave makeup mirror is designed so the virtual image it produces is four times the size of the object when the distance betw

een the object and the mirror is 18.0 cm. What is the radius of curvature of the mirror?

1 answer:

kipiarov [429]3 years ago

7 0

Answer:

Radius of curvature = 48 cm.

Explanation:

Given that

Magnification,m=4

Distance between object and mirror,u=18 cm

We know that magnification of concave mirror given as

m= -v/u

So

4 = -v/18

v= - 72 cm

As we know that

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{f}=\dfrac{1}{18}-\dfrac{1}{72}$

f=24 cm

So r= 2 x f

r=48 cm

Radius of curvature = 48 cm.

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B) The student continues her investigation by loading the spring with different masses. The table shows her result.

liubo4ka [24]

Answer:

1) Id say the dependant is the distance

2) the force is calculated by dividing the mass by 100

Explanation:

the dependant variable is the variable you measure wich is in this case distance.

3 0

3 years ago

For a projectile launched at an angle, are the forces in the vertical direction balanced or unbalanced?

mart [117]

Unbalanced as long as it is moving up or down. Immediately after being fired the vertical force from the launch will be greater than the force of gravity. for an instant at the the beginning of the projectiles decent the force will actually be balanced and there will be no vertical movement. After that the force of gravity is greater.

tl;dr they are unbalanced

7 0

3 years ago

Read 2 more answers

A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a

VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

$r_1 = 45 sin15 \hat i + 45 cos15 \hat j$

$r_1 = 11.64 \hat i + 43.46\hat j$

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

$r_2 = 30 cos15\hat i + 30 sin15 \hat j$

$r_2 = 28.98\hat i + 7.76 \hat j$

now the displacement of the boat is given as

$d = r_2 - r_1$

$d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)$

$d = 17.34 \hat i - 35.7 \hat j$

so the magnitude is given as

$d = \sqrt{17.34^2 + 35.7^2}$

$d = 39.7 km$

4 0

3 years ago

If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can hand

DedPeter [7]

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

$G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$10^2=\dfrac{V_{out}^2}{V_{in}^{2}}$

$\dfrac{V_{out}}{V_{in}^}=10$

$V_{in}=\dfrac{V_{out}}{10}$

$V_{in}=\dfrac{200}{10}$

$V_{in}=20\ mV$

$V_{in}=0.02\ V$

Hence, The largest voltage is 0.02 V.

8 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

4 years ago