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mixer [17]
2 years ago
9

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release

d from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released? A) Its speed will be greatest just after it is released. B) As it moves farther and farther from Q, its speed will decrease. C) As it moves farther and farther from Q, its speed will keep increasing. D) Its acceleration is zero just after it is released. E) As it moves farther and farther from Q, its acceleration will keep increasing.
Physics
1 answer:
andrezito [222]2 years ago
4 0

Explanation:

A positive point charge Q is fixed on a very large horizontal friction less tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move.

As per the law of conservation of energy the change in kinetic energy is equal to the change in potential energy.

The mathematical expression for the conservation of energy is given by :

\dfrac{1}{2}mv^2=-k\dfrac{q_1q_2}{r_2^2}-(-k\dfrac{q_1q_2}{r_1^2})

On solving the above equation, we get the value of v as :

v=\sqrt{\dfrac{2kq_1q_2}{m}(\dfrac{1}{r_1^2}-\dfrac{1}{r_2^2})}

In the above expression, as the term (\dfrac{1}{r_1^2}-\dfrac{1}{r_2^2}) increase, as r_2 increases, the value of \dfrac{1}{r_2} decreases. So, the correct option is (B) "As it moves farther and farther from Q, its speed will increase".

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A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H
ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

M=\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the size of the image

h_o = 2.0 m is the height of the real object (the man)

q=50 mm =0.050 m is the distance of the image from the lens

p = 5.0 m is the distance of the object (the man) from the lens

Solving the formula for h_i, we find

h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm

And the negative sign means the image is inverted.

6 0
2 years ago
13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?
sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of  mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

Q = mc del T

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

Q = 20*C*80

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

5 0
2 years ago
The Olympic record for running the 200m dash is 19.3 seconds. What is the average speed for this record?
Aloiza [94]
I think it’s c. if not i’m sorry!!
3 0
2 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
2 years ago
An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic
Damm [24]

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o  = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

                             y = y_o + v_y,o*t + 0.5*a_y*t^2

                             0 = 10 + 0 + 0.5*a_y*(3.2)^2

                             a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

                             E_p - W_f = E_k

Where,                  E_p = m*g*y_o

                             W_f = m*a_y*(y_i - y_f)      ..... Effects of air resistance

                             E_k = 0.5*m*v_y^2

Hence,                  m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

                             g*(10) - (1.953125)*(10) = 0.5*v_y^2

                             v_y = sqrt (157.1375)

                            v_y = 12.54 m/s

4 0
2 years ago
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