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3 years ago

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According to Ohm’s law, which is stated as I = V ÷ R, which two sentences are true? If the current increases, then the resistanc

e increases. Assume voltage is constant. If the resistance decreases, then the current increases. Assume voltage is constant. If the voltage increases, then the resistance decreases. Assume current is constant. If the voltage increases, then the current increases. Assume resistance is constant. If the voltage increases, then the current decreases. Assume resistance is constant.

2 answers:

stepan [7]3 years ago

5 0

If the voltage increases, then the current increases. Assume resistance is constant. And If the resistance decreases, then the current increases. Assume voltage is constant. Is the correct answer

marta [7]3 years ago

4 0

Explanation :

According to Ohm's law :

V =IR

Where

I is the current flowing

R is the resistance of the wire

V is the potential difference

$I=\dfrac{V}{R}$

If V is constant :

k = IR

$I=\dfrac{k}{R}$

if I increase, then R decrease. So, option (2) is correct.

If R is constant :

$k=\dfrac{V}{I}$

There is an inverse relation between the voltage and the current.

So, If the resistance decreases, then the current increases. Assume voltage is constant.

If the voltage increases, then the current decreases. Assume resistance is constant.

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The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

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Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

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$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

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3 years ago

A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba

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Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

E = Vo / D

E = 600 / 0.3

E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

Fe = E*q

Fe = (2,000 V / m) * ( 3*10^-5 C)

Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

Fe = m*a

a = Fe / m

a = 0.06 / 0.0004

a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

s = C - A

s = ( 0.25 , 12 ) - ( 0.05 , 12 )

s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

vf^2 = vi^2 + 2*a*s

vf^2 = 0 + 2*0.2*150

vf = √60

vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

vf - vi = a*t

t = ( 7.746 - 0 ) / 150

t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

$Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]$

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

$E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\$

- Again, apply the Newton's second law of motion and determine the acceleration (a):

Fe = m*a

a = Fe / m

$a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}$

- Where the acceleration is rate of change of velocity "dv/dt":

$\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\$

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

$c = \frac{A*eo}{d}$

Where, eo = 8.854 * 10^-12 .... permittivity of free space.

$K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\$

- The differential equation turns out ot be:

$\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\$

- Separate the variables the integrate over the interval :

t : ( 0 , t )

v : ( 0 , vf )

Therefore,

$\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )$

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

$vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s$

- The final velocity of particle while charging the capacitor would be:

vf = 7.264 m/s ... slightly less for the fully charged capacitor

7 0

3 years ago

Several different compounds can share the same empirical formula. The Molecular weight for three compounds with the empirical fo

Pie

Molecular formulas:

CH₂O;
C₂H₄O₂;
C₆H₁₂O₆.

<h3>Explanation</h3>

The empirical formula of a compound tells only the ratio between atoms of each element. The empirical formula CH₂O indicates that in this compound,

for each C atom, there are
two H atoms, and
one O atom.

The molecular weight (molar mass) of the molecule depends on how many such sets of atoms in each molecule. The empirical formula doesn't tell anything about that number.

It's possible to <em>add</em> more of those sets of atoms to a molecular formula to increase its molar mass. For every extra set of those atoms added, the molar mass increase by the mass of that set of atoms. The mass of one mole of C atoms, two mole of H atoms, and one mole of O atoms is $12.0 + 2\times 1.0 + 16.0 = 30.0\;\text{g}$ .

CH₂O- 30.0 g/mol;
C₂H₄O₂- 30.0 + 30.0 = 2 × 30.0 = 60.0 g/mol;
C₃H₆O₃- 30.0 + 30.0 + 30.0 = 3 × 30.0 = 90.0 g/mol.

It takes one set of those atoms to achieve a molar mass of 30.0 g/mol. Hence the molecular formula CH₂O.

It takes two sets of those atoms to achieve a molar mass of 60.0 g/mol. Hence the molecular formula C₂H₄O₂.

It takes $\dfrac{180.0}{30.0} = 6$ sets of those atoms to achieve a molar mass of 180.0 g/mol. Hence the molecular formula C₆H₁₂O₆.

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3 years ago

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