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2 years ago

SORRY! COULDN'T EDIT MORE INFO ON MY LAST QUESTION! RE ASKING

Physics

1 answer:

tiny-mole [99]2 years ago

8 0

The position isn't changing ! The object is not moving. It's just sitting there. No matter what 't' you're interested in, the position is exactly the same as it has been since t=0 .

That's (4 divisions) x (2.8 m/div). I'm sure you can calculate the number of meters.

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Please help me promise would mark you brainiest

Triss [41]

B. Please mark me the brilliantest pls

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2 years ago

When we look into the sky every day we get to see the results of all the behaviors of waves. The blue color of the sky results f

mr_godi [17]

Answer:

λ = 0.4 x 10⁻⁶ m = 400 nm

Explanation:

The relationship between frequency, wavelength and speed of an electromagnetic wave is given as follows:

$c = f\lambda$

where,

c = speed of light = 3 x 10⁸ m/s

f = frequency of the light wave = 7.5 x 10¹⁴ Hz

λ = wavelength of the light = ?

Therefore,

$3\ x\ 10^8\ m/s = (7.5\ x\ 10^{14}\ Hz)\lambda\\\\\lambda = \frac{3\ x\ 10^8\ m/s}{7.5\ x\ 10^{14}\ Hz}$

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3 years ago

What is the average velocity of the object between 14 and 22 seconds?

Slav-nsk [51]

-1.1m/s

Explanation:

Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) - U.

4 0

2 years ago

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

Juliette [100K]

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$

3 0

2 years ago

First to answer gets brainliest

Elodia [21]

Answer:

Sodium (K)

Explanation:

6 0

2 years ago