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d1i1m1o1n [39]
3 years ago
6

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is

the acceleration of the ride?
4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2
Physics
2 answers:
NikAS [45]3 years ago
8 0
<span>4.0 m/s2


it's 9 squared divided by 6</span>
yulyashka [42]3 years ago
4 0
The answer is B 4.0 m/s2
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Leokris [45]

They have different amount of neutrons

4 0
3 years ago
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8. If a circuit has a current of 25.4 amps with 12.6
grin007 [14]

The resistance of the circuit is 0.496 \Omega

Explanation:

We can solve the problem by applying Ohm's law, which gives us the relationship between current, potential difference and resistance in a circuit:

V=RI

where

V is the potential difference

R is the resistance

I is the current

For the circuit in this problem,

V = 12.6 V

I = 25.4 A

Re-arranging the equation, we can solve for the resistance:

R=\frac{V}{I}=\frac{12.6}{25.4}=0.496 \Omega

Learn more about current and potential difference here:

brainly.com/question/4438943

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#LearnwithBrainly

8 0
3 years ago
1. Una carga eléctrica de 5*10-6 C genera un campo eléctrico a su alrededor. Calcula la intensidad de este campo a una distancia
Jlenok [28]

Answer:

La intensidad del campo eléctrico es 70312.5 \frac{N}{C}.

Explanation:

La perturbación que crea en torno a ella una carga eléctrica se representa mediante un vector denominado campo eléctrico.

Se dice que un campo eléctrico es uniforme en una región del espacio cuando la intensidad de dicho campo eléctrico es el mismo en todos los puntos de dicha región.

El campo eléctrico E creado por la carga puntual q en un punto cualquiera P se define como:

E=k*\frac{q}{r^{2} }

donde q es la carga creadora del campo, k es la constante electrostática y r es la distancia desde la carga fuente al punto P.

En este caso, los datos son:

  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q= 5*10⁻⁶ C
  • r= 0.8 m

Reemplazando:

E=9*10^{9}\frac{N*m^{2} }{C^{2} }  *\frac{5*10^{-6} C}{(0.8 m)^{2} }

Resolviendo:

E= 70312.5 \frac{N}{C}

<em><u>La intensidad del campo eléctrico es 70312.5 </u></em>\frac{N}{C}<em><u>.</u></em>

5 0
3 years ago
Describe how cell membranes are selectively permeable
attashe74 [19]
<span>Cell membranes are selectively permeable because it allows some things to enter or leave the cell while keeping other things outside or inside the cell.</span>
5 0
3 years ago
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
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