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2 years ago

13

Calculate the speed of a car that traveled 200 kilometers in 3 hours.

2 answers:

nignag [31]2 years ago

8 0

Answer:

The units (km/h) tell you how to do this! 200km/3h = 66.66666666…. BUT technically you only have ONE significant digit: 3 so 66.666… rounded to ONE digit is 70km/h but that is probably not important in this intro class so V = 66.67 or 67 km/h

Kamila [148]2 years ago

7 0

Answer:

66.7km/h

Explanation:

s=d/t

200/3

=66.6666666

round off: 66.7km/h

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Cesar and Jill went to a field to play soccer. As the ball downward toward Jill, Jill used her foot to kick the ball and keep it

Kobotan [32]

Answer:

D

Explanation:

8 0

2 years ago

What is the intensity of 60dB sound?

polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\$

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

$dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2$

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0

2 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

or

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago

When light passes from water into air, the light bends away from the normal. Question 4 options: True False

kykrilka [37]

Answer:

True

Explanation:

8 0

2 years ago

Light enter the container of the liquid at the angle 37° to the normal refraction beam makes an angle of 27° with the normal. Ca

Sliva [168]

Answer: 1.33

Explanation:

We would apply Snell's law which is expressed as

niSinθi = nrSinθr

where

θi = angle of incidence

θr = angle of refraction

ni = index of refraction of the incident medium(air)

nr = index of refraction of the refractive medium(liquid in this case)

From the information given,

ni = 1(index for air)

θi = 37

θr = 27

By substituting these values into the formula, we have

1 * sin37 = nr * sin27

nr = sin37/sin27

nr = 1.33

The index of the liquid is 1.33

6 0

10 months ago