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Dafna1 [17]
3 years ago
15

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

battery, the current through R3 always equals the current through R4. Rcd is always less than or equal to R3. After connecting a and b to a battery, the voltage across R1 always equals the voltage across R2. Rab is always less than or equal to R1.
Physics
1 answer:
Basile [38]3 years ago
5 0

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on R_3 \  and \ R_4, that is the same.

In option 2, this statement is false because Rcd=R_3+R_4 therefore it implies that Rcd is always larger then R_3.

In option 3,  this statement is true because the voltage of R_1 \ and \ R_2 is always equal.

In option 4, this statement is true because Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}is always smaller then 1 therefore,R_1 \times (\frac{R2}{(R1+R2)}) is always equal to R1.

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
Fifty points please help?
Zina [86]

it would be C laminated soda lime glass

6 0
3 years ago
1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam c
SIZIF [17.4K]

Answer:

the power that can be generated by the river is 117.6 MW

Explanation:

Given that;

Volume flow rate of river v = 240 m³/s

Height above the lake surface a h = 50 m

Amount of power can be generated from this river water after the dam is filled = ?

Now the collected water in the dam contains potential energy which is used for the power generation,

hence, total mechanical energy is due to potential energy alone.

E_{mech} = m(gh)

first we determine the mass flow rate of the fluid m

m = p×v

where p is density ( 1000 kg/m³

so we substitute

m = 1000kg/m³ × 240 m³/s

m = 240000 kg/s

so we plug in our values into ( E_{mech} = m(gh) kJ/kg )

E_{mech} = 240000 × 9.8 × 50

E_{mech} = 117600000 W

E_{mech} = 117.6 MW

Therefore, the power that can be generated by the river is 117.6 MW

4 0
3 years ago
A net force of 6.8 N accelerates a 31 kg scooter across a level parking lot. What is the magnitude of the scooter’s acceleration
mrs_skeptik [129]

Answer:

a)0.22 m/s².

Explanation:

Given that

Net force ,F= 6.8 N

mass ,m = 31 kg

From the second law of Newton's

F = m a   ---------------1

Where

F=Net force ,m=mass

a=Acceleration

Now by putting the values in the equation 1

F = m a

6.8 = 31 x a

a=\dfrac{6.8}{31}\ m/s^2

a = 0.219\ m/s^2

a = 0.22\ m/s^2

Therefore the acceleration of the scooter will be 0.22 m/s².

The answer will be "a".

a)0.22 m/s².

3 0
3 years ago
What is the minimum coeffecient of static friction μmin required between the ladder and the ground so that the ladder does not s
mars1129 [50]

Answer:

\mu = \frac{1}{2tan\theta}

Explanation:

let the ladder is of mass "m" and standing at an angle with the ground

So here by horizontal force balance we will have

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by vertical force balance we have

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now by torque balance about contact point on ground we will have

mg(\frac{L}{2}cos\theta) = N_2(L sin\theta)

so we will have

N_2 = \frac{mg}{2tan\theta}

now from first equation we have

\mu (mg) = \frac{mg}{2tan\theta}

\mu = \frac{1}{2tan\theta}

3 0
3 years ago
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