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strojnjashka [21]
3 years ago
9

What is moment and it's type​

Physics
1 answer:
olga2289 [7]3 years ago
8 0

Explanation:

The turning effect of a force is called moment .

There are two types of moment. They are:-

  • clockwise moment
  • anti-clockwise moment
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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
Damm [24]

Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = \dfrac{mv^2}{r}                          

now centrifugal force is balanced by tension

T =\dfrac{mv^2}{r}                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

8 0
3 years ago
Help!!!, combination circuits, Physics
Kaylis [27]

Current and voltage on each resistor:

I_1 = 3.98 A, V_1 = 3.98 V

I_2=0.015 A, V_2 = 0.075 V

I_3 = 0.4 A, V_3 = 0.4 V

I_4 = 0.385 A, V_4 = 0.77 V

I_5 = 0.585 A, V_5 = 1.17 V

I_6 = 3.01 A, V_6 = 6.02 V

I_7 = 0.97 A, V_7 = 4.85 V

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega

This resistor is in series with resistor 4, so:

R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega

This resistor is in parallel with resistor 5, therefore:

R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega

This resistor is in series with resistor 7, so:

R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega

This resistor is in parallel with resistor 6, so:

R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega

Finally, this combination is in series with resistor 1:

R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A

And we can also find the potential difference across resistor 1:

V_1=I_1 R_1=(3.98)(1.0)=3.98 V

This means that the voltage across resistor 6 is

V_6=V-V_1=10-3.98=6.02 V

And so, the current on resistor 6 is

I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

I_7=I-I_6=3.98-3.01=0.97 A

And so the voltage across resistor 7 is

V_7=I_7 R_7=(0.97)(5.0)=4.85 V

The voltage across resistor 5 is

V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V

And so the current is

I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A

The current through resistor 4 is

I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A

And therefore its voltage is

V_4=I_4 R_4 = (0.385)(2.0)=0.77 V

So, the voltage through resistor 3 is

V_3=V_5-V_4=1.17-0.77=0.4 V

And the current is

I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A

Finally, the current through resistor 2 is

I_2=I_4-I_3=0.5-0.385=0.015 A

And so its voltage is

V_2=I_2R_2=(0.015)(5.0)=0.075 V

Learn more about current and voltage:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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4 years ago
Help with 3 please! I will give brainliest!
rusak2 [61]
The answer is point b because vertical velocity is zero at the maximum height
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4 years ago
Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?
miv72 [106K]
The brand of car wax
4 0
3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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