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Leviafan [203]
3 years ago
10

Differentiate between atmospheric pressure and pressure.​

Physics
2 answers:
Dovator [93]3 years ago
8 0

Answer: atmospheric is air by the earth and pressure is just someone or something doing it

Explanation:

RUDIKE [14]3 years ago
5 0

Answer:

Air pressure is the pressure exerted by the air around us while Atmospheric pressure is the pressure exerted by the atmosphere on the earth. Air pressure is measured by tore gauge while atmospheric pressure is measured using mercury barometer.

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A wave with 2.0 m amplitude has a frequency of 500 Hz is travelling at a speed of 200 m/s. What is the wavelength?
Serggg [28]

Answer: 0.4m

Explanation:

Given that:

Amplitude of wave = 2.0 m

Wavelength (λ)= ?

Frequency F = 500Hz

Speed V = 200 m/s

The wavelength is measured in metres, and represented by the symbol λ.

So, apply the formula:

Wavespeed V= Frequency F xwavelength λ

200m/s = 500Hz x λ

λ = 200m/s / 500Hz

λ = 0.4m

Thus, the wavelength is 0.4 metres

3 0
3 years ago
A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×
bagirrra123 [75]
Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
e = 1,6 * 10 ^{-19}C
n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
7 0
3 years ago
The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please
beks73 [17]
I'm pretty sure it's D. The stars don't influence the moon's phases.
6 0
3 years ago
Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

\frac{v}{a_2} = 2t

You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

t_2^2 = 2t_1^2

t_2 = t_1\sqrt{2}\approx 1.14t_1

So t increased by 1.14

7 0
3 years ago
Look at the distance-time graph for a remote-controlled car. What is the speed of the car?
Dafna1 [17]
Speed = distance/time

Speed = 50/5
Speed = 10m/s
7 0
3 years ago
Read 2 more answers
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