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3 years ago

Describe how a change in resistance would affect the current in a circuit.

Physics

1 answer:

Tresset [83]3 years ago

3 0

Explanation:

The relation between potential difference, current and resistance flowing in a circuit is given by using Ohm's law. It can be given by :

V = IR

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

Resistance opposes the flow of electric current in the circuit. It means that, if resistance is more, less current will flow through the circuit.

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As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

fgiga [73]

Answer: $107.8\ J$

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

$\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J$

Initial kinetic energy

$K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J$

Change in kinetic energy is equal to work done by object

$\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J$

5 0

3 years ago

by how much does the speed of a vehicle moving ina straight line change each second when it is accelerating at 2km/h/s? At 4 km/

Svetlanka [38]

Answer:

At 2km/h*s it is changing speed every 2 seconds.

At 4km/h*s it is changing speed every 4 seconds.

At 10km/h*s it is changing speed every 10 seconds.

8 0

2 years ago

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e

Natali5045456 [20]

Answer:

a) $F = 660.576\,N$ , b) $a_{c} = 6.953\,\frac{m}{s^{2}}$ , c) $v \approx 7255.423\,\frac{m}{s}$ , $\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$ , d) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

$F = G\cdot \frac{m\cdot M}{r^{2}}$

$F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}$

$F = 660.576\,N$

b) The centripetal acceleration of the satellite is:

$a_{c} = \frac{660.576\,N}{95\,kg}$

$a_{c} = 6.953\,\frac{m}{s^{2}}$

c) The speed of the satellite is:

$v = \sqrt{a_{c}\cdot R}$

$v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}$

$v \approx 7255.423\,\frac{m}{s}$

Likewise, the angular speed is:

$\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}$

$\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$

d) The period of the satellite's rotation around the Earth is:

$T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)$

$T \approx 1.821\,h$

6 0

3 years ago

Help please! Thanks so much!

qaws [65]

You pretty much just have to convert and be mindful of significant figures.

5 km = 3 miles
0.3 cm = 0.12 in

4 0

3 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

3 years ago