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3 years ago

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

1 answer:

7 0

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

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Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The

Alex73 [517]

Answer:

11.56066 m/s

Explanation:

m = Mass of person

v = Velocity of person = 13.4 m/s

g = Acceleration due to gravity = 9.81 m/s²

v' = Velocity of the person in the second

The kinetic and potential energy will balance each other at the surface

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m$

Height of the cliff is 9.15188 m

Let height of the fall be h' = 2.34 m

$\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s$

The speed of the person is 11.56066 m/s

3 0

3 years ago

Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo

dedylja [7]

Answer:

$\theta = 76.9 degree$

Explanation:

As we know that the resultant of two vectors is given as

$R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta$

here we know that

$R = 87 Lb$

$F_1 = 41 Lb$

$F_2 = 68 Lb$

now we have

$87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta$

$87^2 = 6305 + 5576 cos\theta$

$\theta = cos^{-1}(\frac{1264}{5576})$

$\theta = 76.9 degree$

7 0

3 years ago

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

Llana [10]

Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

a = nb

The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

$h = t \sqrt{1 - \frac{v^2}{c^2} }$

Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

So t = a = nb

and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)

i.e h = b

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$