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jeyben [28]
3 years ago
14

The wave speed in a guitar string of length 57.5 cm is 266 m/s. you pluck the center of the string by pulling it up and letting

go. pulses move in both directions and are reflected off the ends of the string. how many seconds does it take for the pulse to move to the string end and return to the center?
Physics
1 answer:
Anna35 [415]3 years ago
3 0
If I'm correct if you pluck the center of the string creating a pulse off both directions and would be reflected off the ends of the string my calculations would be 10.37sec
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If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to maintain the same leakage curre
harkovskaia [24]
<span>Since P = V x I, a 10% reduction of power would lead to a 10% reduction in the product of voltage and current. What is left is 90% of the original power: .9P = .9(V x I). If we assume that current must be the same, then we can regroup the terms on the right-hand side as follows: .9P = (.9V) x I In this case, voltage is also reduced by 10% (100% - 90% = 10%).</span>
6 0
3 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0
2 years ago
How much potential energy does a 50-N box have when lifted at a height of 1.5M?
nikitadnepr [17]

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (<em>mg</em>) of the box is given as the unit is <em>Newton</em>, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)


In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5  = <em>75 J (Option A)</em>

3 0
3 years ago
When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4
vitfil [10]
Just divide the two numbers with each other.
I mean 13/1.4=9.2857...
8 0
2 years ago
HELPPP ME PLEASE!!!!
andriy [413]

Answer:

c

Explanation:

3 0
2 years ago
Read 2 more answers
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