All categories

2 years ago

Conclusions _____.

Physics

1 answer:

PIT_PIT [208]2 years ago

5 0

Answer:

hjk

Explanation:

hjkhjk

You might be interested in

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

$I=2.50\cdot 10^4 A$ is the current

$t=40.0 \mu s = 40.0\cdot 10^{-6} s$ is the time interval

Solving the formula for q, we find the amount of charge transferred:

$q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C$

6 0

2 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

A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate

slava [35]

The refractive index of water is $n=1.33$ . This means that the speed of the light in the water is:
$v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s$

The relationship between frequency f and wavelength $\lambda$ of a wave is given by:
$\lambda= \frac{v}{f}$
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water $\lambda_w$ to that in air $\lambda$ as
$\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}$
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using $\lambda=400 nm$ , we find
$\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm$
which is the wavelength of light in water.

5 0

2 years ago

A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (

kogti [31]

Answer: See below

Explanation:

Given:

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

The free body diagram is shown in figure attached below

a) The normal force is equal to the cos component of the weight of the car.

$\begin{aligned}&f=m g \cos \theta \\&f=1700 \times 9.81 \times \cos 15 \\&f=16108.74 \mathrm{~N}\end{aligned}$