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3 years ago

Which statement about atmospheric pressure near earth's surface is true

Physics

1 answer:

Artemon [7]3 years ago

7 0

The correct answer is it decreases as someone moves away from earth's surface.

The atmospheric pressure decreases when you go near the surface of the earth. Atmospheric pressure can also be referred to as barometric pressure.

This is the pressure which is being found in the earth's atmosphere. It is approximated by pressure of hydrostatic which is known to be caused by weight of air where the measurement points are above.

When elevation increases, there is decrease in atmospheric pressure.

Force per unit are is measured by pressure.

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What is the following correct way to write 2,330,000 In a scientific notation

NISA [10]

Answer:

2.33 × 10^6

hope this helps.

8 0

3 years ago

Read 2 more answers

The exhaust fan on a typical kitchen stove pulls 600 CFM (cubic feet per minute) through the filter. Given that 1.00 in. = 2.54

zysi [14]

Answer:

$0.283\ m^3/s$

Explanation:

Given that,

1.00 in = 2.54 cm

We know that,

1 feet = 12 inches

$1\ feet =12\times0.0254$

$1\ feet =0.3048\ m$

We calculate the number

We need to converted $ft^3/min$ to $m^3/sec$

The number required is

$600ft^3/min = \dfrac{600\times(0.3048)^3}{60}$

$600ft^3/min=0.283\ m^3/s$

Hence, This is the required solution.

4 0

3 years ago

Bola bermassa 200 gram dilempar

ikadub [295]

Answer:

0.4

Explanation:

$\frac{1}{2} mv ^{2}$

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

8 0

3 years ago

8_murik_8 [283]

Answer: 2.61 s

Explanation:

We are given the following data:

$s=5 ft$ is the initial height of the object

$v=40 ft/s$ is the initial height of the object

In addition, the motion of the object is given by:

$h=-16t^{2}+ vt +s$ (1)

Where:

$h=0 ft$ is the final height of the object

$t$ is the time the object is in the air before hitting the ground

Rewritting (1) with the given data:

$0=-16t^{2}+ 40t +5$ (2)

Solving the quadratic equation with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ , where $a=-16$ , $b=40$ , $c=5$ and choosing the positive value of time:

$t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)}$ (3)

$t=2.61 s$ (4) This is the time

5 0

3 years ago

An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A

melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

$\omega = \frac{\theta}{t}$

Where θ: is the angular displacement = π (since it moves halfway through a complete circle)

We have:

$t = \frac{\theta}{\omega} = \frac{\theta}{v/r}$

Where:

v: is the tangential speed

r: is the radius

The radius can be found equaling the magnetic force with the centripetal force:

$qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}$

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

$t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s$

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!

4 0

3 years ago