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ElenaW [278]
3 years ago
7

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

e on the y-axis. Find an expression for (Fnet)x, the x-component of the net force on q. (Give your answer in terms of Q, q, a, y and constant K.)
Physics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

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I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0
3 years ago
A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

8 0
3 years ago
What is the trend noticed with the inner planets
lutik1710 [3]
The inner planets are closer to the Sun and are smaller and rockier. ... The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars.Apr 23, 2014
7 0
3 years ago
Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat
Nataly_w [17]

Answer:

\omega_{f} = 1000000\,\frac{rev}{day}

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}

The final angular speed is:

\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}

\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}

\omega_{f} = 1000000\,\frac{rev}{day}

3 0
2 years ago
Which of the following is a way earth’s water (hydrosphere) is directly affected by pollution?
ASHA 777 [7]

The toxic runoff from the farms will directly enters the water bodies of the earth through the drainage system.

The emissions from the cars will go into atmosphere and will react with water and then it will enter the earth's water. So, it is not directly affecting the earth's water.

The dumping trash on roadsides do not enter the earth's water. It can enter the earth's water through rains.

The production of oxygen by trees is not pollution.

Hence, option (A) is correct.

6 0
2 years ago
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