Ask question

Ask question

All categories

3 years ago

13

Light travels in... A) electromagnetic waves B) electrons C) photons D) a and c

1 answer:

julsineya [31]3 years ago

4 0

Answer:

Light travels in electromagnetic waves.

You might be interested in

The vaule of the g constant (the acceleration of all objects due to gravity)on earth is

True [87]

The answer is: " $\frac{9.8 m}{s^2}$ " .
_____________________________________________________

6 0

3 years ago

A 2-Newton upward net force is being applied to a 10-kg object. What is the magnitude of the upward

Colt1911 [192]

Acceleration in m/s^2 = 2/10 = 0.2 m/s^2

7 0

3 years ago

An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa

Tatiana [17]

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm$

5 0

3 years ago

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

3 years ago

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

<h3>200 Hz</h3>

Hope this helps you

7 0

3 years ago