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Dmitrij [34]
3 years ago
6

A particle leaves the origin with an initial velocity v = (3.00 i hat) m/s and a constant acceleration a = (-3.00 i hat - 1.400

j hat) m/s2
(a) What is the velocity of the particle when it reaches its maximum x coordinate?

m/s i hat + m/s j hat + m/s k hat

(b) What is the position vector of the particle at this time?

m i hat + m j hat + m k hat
Physics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

a)  —0.5 j m/s  

b) 4.5 i + 2.25 j m

Explanation:

<u>Givens:</u>

v_0 =3.00 i m/s

a= (-3 i — 1.400 j ) m/s^2  

The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :

<em>v_x = v_0 + at = 0  </em>

(3.00 i m/s) + (-3 i m/s^2)t=0

t = (3 m/s)/-3 i m/s^2

t = -1 s

Therefore the particle reaches the maximum x-coordinate at time t = 1 s.  

Part a The velocity-of course- is all in the y-direction,therefore:

v_y =v_0+ at

We have that v_0 = 0 in the y-direction.

v_y = (-0.5 j m/s^2)(1 s)

      = —0.5 j m/s  

Part b: While the position of the particle at t = 1 s is given by:

r=r_0+v_0*t+1/2*a*t^2

Where r_0 = 0 since the particle started from the origin.

Its position at t = 1 s is then given by :

r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2

 =4.5 i + 2.25 j m

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An RLC circuit is used in a radio to tune into the radio lagos fm Station broadcasting at 93.5Hz. The resistance is 15ohms and t
bazaltina [42]

The characteristics of the RLC circuit allow to find the result for the capacitance at a resonance of 93.5 Hz is:

  • Capacitance is C = 1.8 10⁻⁶ F

A series RLC circuit reaches the maximum signal for a specific frequency, called the resonance frequency, this value depends on the impedance of the circuit.

            Z^2 = R^2 + ( wL - \frac{1}{wC} )^2  

Where Z is the impedance of the circuit, R the resistance, L the inductance, C the capacitance and w the angular velocity. The negative sign is due to the fact that the current in the capacitor and the inductor are out of phase.

In the case of resonance, the impedance term completes the circuit as a resistive system.

           wL - \frac{1}{wC} = 0  \\w^2 = \frac{1}{LC}  

           

Indicate that the inductance L = 1.6 H and the frequency f = 93.5 Hz.

Angular velocity and frequency are related.

         

         w = 2π f

           

Let's  substitute.

          C = \frac{1}{L ( 2 \pi f)^2 }  

 

Let's calculate.

         C = \frac{1}{1.6 \ ( 2\pi \ 93.5)^2}  

         C = 1.8 10⁻⁶ F

In conclusion with the characteristics of the RLC circuits we can find the result for the capacitance at a 93.5 Hz resonance is:

  • Capacitance is C = 1.8 10⁻⁶ F

Learn more about serial RLC circuits here: brainly.com/question/15595203

4 0
2 years ago
When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
Oxana [17]

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

m=5\cdot 10^{-3}g

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

1 g : 1000 mg = 0.005 g : x

and we find

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

4 0
3 years ago
If a wave has speed of 235 m/s with a wavelength of 3 m, what is the frequency of the wave?
maria [59]
I think it’s 25 but I don’t know
5 0
2 years ago
Alex want to test factors that affect the reaction rate of sugar in water
miv72 [106K]
This aint even a question

7 0
2 years ago
a 13-gram bullet, moving at 270 m/s, penetrates a 2 kg block of wood and emerges at a speed of 130 m/s. if teh block sits one a
saveliy_v [14]

Answer:

1.52m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute the given values into the formula

0.013(270)+2(130) = (270+130)v

3.51+260 = 400v

263.51 = 400v

v = 400/263.51

v = 1.52m/s

Hence the velocity after the bullet emerges is 1.52m/s

6 0
3 years ago
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