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Keith_Richards [23]
3 years ago
7

Does anyone understand circuits?

Physics
1 answer:
Alina [70]3 years ago
5 0
Hope this helps you.

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Please help ASAP WILL GIVE BRAINLIEST
Westkost [7]

Answer:

D

Explanation: It makes the most sense. Plz mark brainliest

5 0
3 years ago
Read 2 more answers
Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.
lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have   .21   ( is that down? I will assume it is)

      Counterclockwise moments :

        .21 * 40     +  1.0 * 20    

     Clockwise moments :

        .5 * 20     +     F * 45

these moments must equal each other

.21*40 + 1 *20   =  .5 * 20 + F * 45

   F = .409 N

7 0
2 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0
3 years ago
A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel
lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

\epsilon = -NA\frac{dB}{dt}

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}

\epsilon = 0.212V

\epsilon = 0.21V

Therefore the correct answer is A.

6 0
3 years ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
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