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4 years ago

Plucking a guitar string is a physical change; however, the process produces sound. Explain. plz heip

1 answer:

7 0

The vibration of the string moves the surrounding air molecules. The air molecules move in the form of a wave, traveling a distance dependent upon the magnitude of the vibration.

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A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

7 0

3 years ago

What do you meant by varnier constant?

Soloha48 [4]

Answer:

the ratio of the smallest division of main scale to the number of divisions of the vernier scale.

Explanation:

difference between the value of one main scale division and one vernier scale division

3 0

3 years ago

A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves

Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

$U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\$

So, the work done to move the chargue is:

$W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J$

The work is positive since the potential energy in 1 is greater than 2.

5 0

4 years ago

Help yet again :) A hockey player is skating on the ice at 15km/h. He shoots the puck at 138 km/h according to a radar gun on th

IceJOKER [234]

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

7 0

3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$