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2 years ago

15

A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v

ertically downwards.
a) Draw vectors of forces acting on the sides of the frame.
b) calculate the values of these forces and the resultant force acting on the frame.
c) calculate the torque of the frame.

1 answer:

IRISSAK [1]2 years ago

6 0

The solution to this task is f00king Dis.

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A baseball bat hits a baseball with a force of 100 newtons. What is the force and its direction exerted by the ball on the bat?

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Answer:

The force exerted by the ball on the bat has a magnitude of 100 N and its direction is exactly opposite to that of the force exerted by the bat on the ball.

Explanation:

Recall that Newton's third law tells us that : "For every action, there is an equal and opposite reaction."

Therefore if the bat acts on the ball with a force of 100 N, the ball acts on the bat with a similar magnitude of force (100 N) but direction opposite to the original force.

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For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr

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Answer:

L= 2 mH

$C=1.26\times 10^{-7}\ F$

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

$E=\dfrac{1}{2}LI^2$

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

$10^{-5}=\dfrac{1}{2}\times L\times 0.1^2$

$L=\dfrac{2\times 10^{-5}}{0.1^2}\ H$

L=0.002 H

L= 2 mH

We know that frequency f is given as

$2\pi f=\dfrac{1}{\sqrt{LC}}$

C=Capacitance , f=frequency ,L=Inductance

Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

$0.002\times C=0.0000159^2$

$C=\dfrac{0.0000159^2}{0.002}\ F$

$C=1.26\times 10^{-7}\ F$

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

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3 years ago

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be

Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

$v_1=\dfrac{72.42 }{2.5}=28.96\ km/h$

Average speed of Prav,

$v_2=\dfrac{119}{1.25}=95.2\ km/h$

Difference in average speed of Ajoba and Prav is :

$v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h$

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

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Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a

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Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

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3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

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Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

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Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

So

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

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3 years ago