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EastWind [94]
3 years ago
15

What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-

directed electric field of magnitude 670 n/c ?
Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg  

Take downward direction as positive.  

Fg = m g  

E = 670 N/C  

Fe = q E  

Fe + Fg = 0  

q E + m g = 0  

q = -m g/E

   = -1.49 × 10^-3 × 9.81/670

    = -2.18 × 10^-5 C  

=  -2.18 × 10^-5 C

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Eddi Din [679]

a)E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) <em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude  is 'A'

\omega=\sqrt{\frac{k}{m}} is the angular frequency of the motion

t is the time

\phi is the phase (we can take \phi=0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K=\frac{1}{2}mv^2=0

E=\frac{1}{2}kA^2\\ -->(1)

<em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

c)v_{max}=\omega A<u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E=\frac{1}{2}mv_{max}^2

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2

and solving for v_{max}we find an expression for the maximum speed:

v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A

<h2><u></u>v_{max}=\omega A<u></u></h2>
4 0
3 years ago
Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t
yawa3891 [41]

Answer:

a) x=2*10^{9} m and t=8.35 s

b) t = γt', so it is 8.35 s.

Explanation:

a) The equation of Lorentz transformations is given by:

x=\gamma(x'+ut')  

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

  • x' = 0
  • t' = 5.00 s
  • u =0.800 c, c is the speed of light 3*10⁸ m/s

\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}

\gamma=1.67

x=1.67(0+0.800c*5.00)

x=2*10^{9} m

Now, to find t we apply the same analysis:

t=\gamma(t'+\frac{ux'}{c^{2}})                        

but as x'=0 we just have:

t=\gamma(t')

t=1.67*5.00=8.35 s

b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.

I hope it helps you!

5 0
3 years ago
Read 2 more answers
Calculate the magnitude of the gravitational force between Goku with a mass of 62 kg and King Kai’s planet with a mass of 1.458x
Brilliant_brown [7]

Answer:

6227.866 N

Explanation:

F = G . m(goku) . m(planet) / d²

F = 6.674 x 10-¹¹ x 62 x 1.458 . 10¹⁵ / 31²

F = 6227.866 N

7 0
2 years ago
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho
Andrews [41]

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}

\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)

[tex]\overrightarrow{\tau  }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

3 0
3 years ago
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