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3 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

1 answer:

4 0

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>

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In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor

VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

$I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596$ kg-m2

$I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2$

by perpendicular axis theorem

$I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2$

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3 years ago

Which statement accurately describes the molecules of a gas?

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3 years ago

The leader of a bicycle race is traveling with a constant velocity of +12.60 m/s and is 11.4 m ahead of the second-place cyclist

Svetllana [295]

Let the time be t.

so,In time t , distance travelled by 1st cyclist = 12.6 t

distance travelled by 2nd cyclist = 9.2t + 0.5 (1.5) t^2

Now, cyclist 1st is already 11.4 m ahead of 2nd cyclist.

so, 9.2t + 0.5 (1.5) t^2 = 11.4 + 12.6t

find t :

t = 6.77 sec

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3 years ago

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Answer:

Explanation:

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3 years ago

A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 89 m/s. At t = 0, t

earnstyle [38]

Answer:

The answer is below

Explanation:

Given that:

y = transverse displacement = 4.2 cm = 0.042 m at x = 0 and t = 0.

Speed = v = 89 m/s, maximum transverse speed of the string particle = $u_m$ = 16 m/s.

ω = $u_m$ / $y_m$ = 16 / 0.42 = 380.95 rad/s

a) Frequency = ω/2π = 380.95 / 2π = 60.63 Hz

b) Wavelength (λ) = speed / frequency

λ = v / f = 89/63.66= 1.468 m

c) Using the wave equation:

$y=y_msin(kx \pm wt \pm \phi)\\y=0.042,t=0,x=0\\\\Hence\\y_m=0.042\ m$

d) Wave number k is given by:

k = 2π / λ = 2π / 1.468 = 4.28 rad/s

e) The angular velocity is given by:

ω = $u_m$ / $y_m$ = 16 / 0.42 = 380.95 rad/s

f) Using the wave equation:

$y=y_msin(kx \pm wt \pm \phi)\\\\y=0.042,t=0,x=0,y_m=0.042\\\\Hence\\0.042=0.042sin(4.28(0)\pm 380.95(0)\pm \phi)\\\\sin\phi=1\\\\\phi=\frac{\pi}{2} \\\\y=0.042sin(4.28x\pm 380.95t\pm \frac{\pi}{2})$

g) Since the wave is in the positive x direction, hence ω is negative

4 0

3 years ago