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3 years ago

Why is it important to consider experimental error in all the empirical results presented?

1 answer:

3 0

Because when your running any experiment there will always be an experimental error so always be ready for it so that you can correct that error in your experiment

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The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?

yuradex [85]

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, $L=9.4\ mH=9.4\times 10^{-3}\ H$

Resistance, R = 150 ohms

Capacitance, $C=1.9\ \mu F=1.9\times 10^{-6}\ C$

At resonance, the capacitive reactance is equal to the inductive reactance such that,

$X_C=X_L$

$2\pi f_o L=\dfrac{1}{2\pi f_oC}$

f is the resonant frequency of this circuit

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

$f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}$

$f_o=1190.91\ Hz$

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

4 0

3 years ago

Why do scientists believe that light is made of streams of particles?

irina [24]

I just need points to ask my own question.

5 0

3 years ago

Read 2 more answers

two teams are playing tug of war. team a pulls to the right with a force of 450n .team b pulls to the left with a force of 415 n

Alex_Xolod [135]

Explanation:

It is given that, two teams are playing tug of war.

Force applied by Team A, $F_A=450\ N$

Force applied by Team B, $F_B=415\ N$

We need to find the net force acting on the rope. It is equal to :

$F_{net}=F_A-F_B$

$F_{net}=450-415$

$F_{net}=35\ N$

So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.

4 0

3 years ago

The Pacific Giant Kelp grows at a rate of 18 in/day. How many centimeters per hour is this?

Airida [17]

For this case we must do a conversion. By definition we have to:

1 inch equals 2.54 centimeters

1 day equals 24 hours.

So, we have: $18 \frac {in} {day} * \frac {1} {24} \frac {day} {h} * \frac {2.54} {1} \frac {cm} {in} =$

Canceling units we have:

$\frac {18 * 1 * 2.54} {24 * 1} \frac {cm} {h} =\\\frac {45.72} {24}\frac {cm} {h} =$

$1,905 \frac {cm} {h}$

Answer:

$1,905 \frac {cm} {h}$

5 0

3 years ago

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

defon

This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:

Ff = UsN 

where Us is the coefficient of static friction and N is the normal force. 

In order to get the crate moving you must first apply enough force to overcome the static friction: 

Fapplied = Ff 

Since Fapplied = 43 Newtons: 

Fapplied = Ff = 43 = UsN 

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2

4 0

3 years ago

Read 2 more answers