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1 year ago

If the resistance in a circuit connected to a constant current is halved, how is the voltage in the circuit affected?.

Physics

2 answers:

Archy [21]1 year ago

6 0

Answer:

The voltage remains constant.

Vadim26 [7]1 year ago

3 0

Answer:

The voltage in the circuit remains same. However the current drawn by the circuit will increase.

Explanation:

According to Ohm's Law:

V=IR

If we halved the resistance let:

R'=R/2

then

I'=2I in order to kept the Voltage same. Also the resistance is the opposition to the current. So, if resistance will halved then obviously current will double

So when we put these value in to Ohm's Law the results is:

V'=I'R'

V'=2I(R/2)

V'=IR which is equal to first voltage. Hence,

V'=V

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Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.17 m on a

Nastasia [14]

Answer: The last part of the question has some details missing which is ; (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) molecule v=482 m/s molecule momentum=2.56 x 10^(-23)

Explanation:

The momentum of the molecule is 2.56 x 10^(-23) .
Particle hits the wall and bounces.
Momentum is reversed. Change in momentum = impulse
This is Force x time.
Momentum change happens at a wall after each trip.

time required = distance /speed

= 0.17 X 2/(482 m/s)

Average force = impulse / time

= 2 x 482 x 2.56 x 10^(-23) / (0.17 x 2)

= 7.76 x 10^20N, is the average force the molecule exerts on one of the walls of the container.

3 0

4 years ago

Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your

forsale [732]

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

$\\ \sf\longmapsto F_L=\mu sN$

u s is coefficient of static friction and N is normal reaction

$\\ \sf\longmapsto F_L=\mu smg$

As N=mg

8 0

3 years ago

Read 2 more answers

A uniform solid sphere rolls down an incline. A) What must be the incline angle (deg) if the linear acceleration of the center o

Nonamiya [84]

Answer:

a) θ = 12.12°

b) equal to 0.21g

Explanation:

Solution:-

Declare variables:

- The mass of solid sphere, m

- The inclination angle, θ

- The linear acceleration a down the slope of the solid sphere is a = 0.21g

Where, g: The gravitational acceleration constant.

- The component of weight of solid sphere directed down the slope is given by:

Ws = m*g*sin ( θ )

- Apply Newton's second law of motion down the slope, state:

F_net = m*a

- The only net force acting on the solid sphere is the Weight. So, the equation of motion in the coordinate axis ( down the slope ).

Ws = m*a

m*g*sin ( θ ) = m*0.21*g

- Solve for inclination angle ( θ ):

sin ( θ ) = 0.21

θ = arcsin ( 0.21 )

θ = 12.12°

- If a friction-less block of mass ( m ) moves down the same slope then block has weight component down the slope as:

Wb = m*g*sin ( θ )

- Apply Newton's second law of motion down the slope, state:

F_net = m*a

- The only net force acting on the solid sphere is the Weight. So, the equation of motion in the coordinate axis ( down the slope ).

Ws = m*a

m*g*sin ( θ ) = m*a

- Solve for linear acceleration ( a ):

g*sin ( θ ) = a

a = sin ( 12.12 ) * g

a = 0.21g

Answer: The acceleration is independent of mass and only depends on the inclination angle θ.

7 0

4 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

4 years ago

A child riding a bicycle at 5 meters per seconds accelerates at 2.0 meters per second for 4.0 seconds. What is the child's speed

slega [8]

Explanation:

Using Kinematics,

We have v = u + at.

=> v = (5m/s) + (2.0m/s²)(4.0s)

=> v = 13m/s.

Hence the child's speed is now 13m/s.

5 0

3 years ago