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kodGreya [7K]
4 years ago
7

??????????????????????

Physics
2 answers:
Galina-37 [17]4 years ago
4 0
Electrons are found in shells or orbitals that surround the nucleus of an atom
olga55 [171]4 years ago
3 0
All electrons are found in the outer shells orbiting the nucleus. Hope this helps!
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Looking at group 2 on the periodic table which of these elements probably has the lowest ionic radius
lakkis [162]
I think the answer is Helium.

Please tell me if I'm wrong.
7 0
3 years ago
Why does Farm Bureau and other advocacy organization oppose any mandated labeling of biotech crops?
Step2247 [10]

Answer:

I’m. Nog sure

Explanation:

3 0
3 years ago
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi
zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate\\\rho = Density\\V = Velocity

values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3V_1 = 40m/s

A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s

PART A) Applying the flow equation

\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2

7 0
3 years ago
If the value of the resistor r2 were doubled, how would the value of the resistor r3 have to change in order to keep the current
Reika [66]

This is a Wheatstone bridge, and the ratio of R2 to R1 equals the ratio of Rx to R3. As a result, if R2 is increased, R3 should be reduced by a factor of two.

<h3>Explain Wheatstone bridge?</h3>

A Wheatstone bridge is a type of electrical circuit that is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one of which contains the unknown component.

The Wheatstone bridge circuit can be used to compare an unknown resistance RX to others of known value, such as R1 and R2, which have constant values and R3 which can be variable.

If we connected a voltmeter, ammeter, or galvanometer between points C and D, and then changed resistor R3 until the meters read zero, the two arms would be balanced, and the value of RX (substituting R4) would be known as indicated.

To learn more about Wheatstone bridge refer to :

brainly.com/question/15225070

#SPJ4

5 0
1 year ago
Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t
ziro4ka [17]

Answer:

(a). The acceleration is 8.3 rad/s².

(b). The time is 3.0 sec.

Explanation:

Given that,

Rotation = 20.0 times

Time = 5.00 sec

We need to calculate the angular frequency

Using formula of angular frequency

\omega_{i}=20\times\dfrac{2\pi}{T}

put the value into the formula

\omega_{i}=20\times\dfrac{2\pi}{5.00}

\omega_{i}=8.00\pi

We need to calculate the angular displacement

Using formula of  angular displacement

\theta=6\times2\pi=12\pi

We need to calculate the angular acceleration

Using equation of angular motion

\omega_{f}^2-\omega_{i}^2=2\alpha\times\theta

0-64\pi^2=2\times12\pi\times\alpha

\alpha=-\dfrac{64\pi^2}{24\pi}

\alpha=-8.3\ rad/s^2

Negative sign shows the opposite direction of the motion.

The acceleration is 8.3 rad/s².

We need to calculate the time

Using equation of angular motion

\omega_{f}-\omega_{i}=\alpha\times t

0-8\pi=-8.3t

t=\dfrac{8\pi}{8.3}

t=3.0\ sec

The time is 3.0 sec.

Hence, (a). The acceleration is 8.3 rad/s².

(b). The time is 3.0 sec.

4 0
3 years ago
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