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2 years ago

What is a non example of linear relationship

Physics

1 answer:

xxTIMURxx [149]2 years ago

4 0

Neither the speed nor the distance of a falling object is linearly related to time.

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51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

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What kind of friction exists between solid objects moving in water?

eduard

Fluid Friction exists when it is acted upon an object when in fluid.

6 0

3 years ago

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

CaHeK987 [17]

Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N

4 0

3 years ago

For questions 1-10 fill in the blank with the letter of the description that best matches the term

tigry1 [53]

B-Pitcher C-Catcher H-Strike I-Umpire G-Strike Zone E- Foul Ball F- Ball J- Pick-off D-Error A- Shortstop. I think (Sorry for them being out of order. I had to break them down)

4 0

3 years ago

Read 2 more answers