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4 years ago

Personality tests are one of the most accurate ways to let you know exactly who you are

1 answer:

8 0

If this is a True/False question the answer is False. the most accurate ways to let you know exactly who you are is to... well, be exactly who you are! That's all I can really say here. Personality tests may be on target sometimes but only you can know who you truly are, you know?

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Burning oil and coal adds to the atmosphere.

True [87]

Answer:

carbon dioxide

Explanation:

8 0

4 years ago

What is the sig fig of 0.00324?

Nonamiya [84]

This number has 3 sig figs.

5 0

3 years ago

Compare the atmospheric pressure of two places 500m and 2 km respectively above sea-level. Give reason for your answer.

Lera25 [3.4K]

Answer:

Explanation:

The average pressure at mean sea-level (MSL) in the International Standard Atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury. Pressure (p), mass (m), and the acceleration due to gravity (g), are related by P = F/A = (m*g)/A, where A is surface area.

3 0

3 years ago

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?

Vedmedyk [2.9K]

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h3>Given:-</h3>

Velocity (V) = 40 m/s

Wavelength $(\lambda)$ = 80 cm = 0.8 m

The frequency (F) of the wave.

<h2>Solution:-</h2>

We know,

$\bf V \: = \: F \: × \: \lambda$

40 = F × 0.8

F = $\frac{40}{0.8}$

F = 50

<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>

7 0

3 years ago

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Elan Coil [88]

Answer:

$a=0.284\ m/s^2$

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$

So, the acceleration experienced by the occupants of the spaceship is $0.284\ m/s^2$ .

5 0

3 years ago