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2 years ago

How u do this atomic mass

Physics

2 answers:

bezimeni [28]2 years ago

6 0

Vas happenin!!

To find atomic mass you you add the protons and the neutrons

Hydrogen-
Atomic number: 1
Atomic mass: 1.008
Number of protons: 1
Number of electrons: 1
Number of neutrons: 1

Helium-
Atomic number: 2
Atomic mass: 4.0026
Number of protons: 2
Number of neutrons: 2
Number of electrons: 2

Hope this helps

-Zayn Malik

Eddi Din [679]2 years ago

5 0

Explanation:

Hydrogen

Atomic Number=1

Atomic Mass= 1

number of protons= 1

number of electrons= 1

number of neutrons= 0

Helium

atomic number=2

atomic mass=2

number of protons=2

number of electrons=2

number of neutrons=2

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Which phenomenon occurs when a wave encounters a non–transmitting barrier

SpyIntel [72]

Answer:

Absorption

Explanation:

A non-transmitting barrier would not allow a wave to go through. When a wave is unable to pass through a barrier, it is not transmitted and can get absorbed or reflected back. The wave can also try to go round the barrier.

Most likely, the wave gets absorbed by the barrier and it stops it.

6 0

3 years ago

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I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m

vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

$t=\dfrac{56}{29}\ s$

Distance covered in 0.42 seconds

$0.42\times 20=8.4\ m$

From equation of linear motion

$s=ut+\frac{1}{2}at^2$

$8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2$

The minimum acceleration is 6.46393559312 m/s²

5 0

3 years ago

An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes

Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = $\frac{1}{2}$ ( 60 cm - 10 cm )

A = $\frac{1}{2}$ × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed $V_{max}$ = ωA

and ω = 2π/T

so maximum speed $V_{max}$ = $\frac{2\pi }{T}$ A

so we substitute

so maximum speed $V_{max}$ = $\frac{2\pi }{3.3}$ × 0.25 m

so maximum speed $V_{max}$ = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

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2 years ago

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2 years ago

Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi

VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

$t = \sqrt{0.388} = 0.62 s$

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0

2 years ago