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3 years ago

A 1.5m wire carries a 3 A current when a potential difference of 86 V is applied. What is the resistance of the wire?

1 answer:

8 0

We know, R = V / I
Here, V = 86 V
I = 3 A

Substitute their values,
R = 86 / 3
R = 28.67 Ohm

In short, Your Answer would be 28.67 Ohms

Hope this helps!

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f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

3 years ago

A car has an initial velocity of 20m/s and an average velocity of 30m/s.

Luba_88 [7]

Answer:

call 7194936846 and ill help u out

Explanation:

3 0

2 years ago

How do I solve this step by step? I’m really confused

LekaFEV [45]

Step-#1:

Ignore the wire on the right.

Find the strength and direction of the magnetic field at P,

caused by the wire on the left, 0.04m away, carrying 5.0A

of current upward.

Write it down.

Step #2:

Now, ignore the wire on the left.

Find the strength and direction of the magnetic field at P,

caused by the wire on the right, 0.04m away, carrying 8.0A

of current downward.

Write it down.

Step #3:

Take the two sets of magnitude and direction that you wrote down

and ADD them.

The total magnetic field at P is the SUM of (the field due to the left wire)

PLUS (the field due to the right wire).

So just calculate them separately, then addum up.

4 0

3 years ago

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati

Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π $\sqrt{\frac{l}{g} }$

angular frequency ω = 2π /T

= $\sqrt{\frac{g}{l} }$

ω₁ /ω₂ = $\sqrt{\frac{g_1}{g_2} }$

Putting the values

ω₁ = ω , ω₂ = ω / 4

ω₁ /ω₂ = 4

4 = $\sqrt{\frac{g}{g_2} }$

g₂ = g / 16

option d is correct.

6 0

3 years ago

A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .

swat32

From the calculations, the power expended is 43650 W.

<h3>What is the power expended?</h3>

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v = 95 km/h or 26.4 m/s

t = 6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work = 3298 N * 90 m = 296820 J

Power = 296820 J/ 6.8 s

= 43650 W

Learn more about power expended:brainly.com/question/11579192

#SPJ1

5 0

2 years ago