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3 years ago

6. What do the key results indicate?

Physics

2 answers:

slega [8]3 years ago

7 0

the key result indicate that how you will achieve the your goal

yuradex [85]3 years ago

3 0

Answer:

The key results explains how you achieve a certain goal. They have to be quantitative, but always measurable in terms of their progress.

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A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter

yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

where Rint = r and RL = r
Replacing these values in I₁, we have:

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

2 years ago

A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?

Over [174]

Answer:

$\sqrt{ ({40}^{2} } + {50}^{2} ) = \sqrt{(1600 + 2500)} = \sqrt{4100} = 64.03$

6 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

2 years ago

If a machine will do 50 J of work in 35 seconds how much power did it produce?

grandymaker [24]

Power is the amount of work done over a period of time. If you will put that into an equation, the formula of power will be:

P = W/t
Where:
P = power
W=work
t = time

Your problem already provides you with work and time so all you need to do is divide:

P = W/t
P = 50J/30s
P = 1.67 W

6 0

2 years ago