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3 years ago

A convex lens has a focal length of 0.33 m. The object distance is 0.7 m. What is the image distance?

Physics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer:

Explanation:

1/v - 1/u = 1/f

given, f = 0.3 m, u = -0.4m

so, 1/v - 1/-0.4 = 1/0.3

or, 1/v = 1/0.3 - 1/0.4 = 1/1.2

v = 1.2 m

now, differentiating 1/v - 1/u = 1/f with respect to t,

-1/v² dv/dt + 1/u² du/dt = 0

or, dv/dt = (v/u)² du/dt

putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m

so, dv/dt = (1.2/-0.4)² × 0.01

= 0.09 m/s

hence, speed of image with respect to lens is 0.09 m/s .

from formula of magnification

magnification, m = v/u

differentiating with respect to time both sides,

dm/dt = (u dv/dt - vdu/dt)/u²

= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²

= (-0.036 - 0.012)/0.16

= -0.048/0.16

= -0.3 m/s

hence, magnitude of rate of change of lateral magnification is 0.3 m/s

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Does the speedometer of a car measure speed or velocity? Explain.

deff fn [24]

Car speedometer only measures speed and doesn't give any information about direction. So yes to speed, no to velocity. ... Therefore the object CANNOT have a varying speed if its velocity is constant.

6 0

3 years ago

Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann

chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

$v_{mp}=\sqrt{\frac{2K_bT}{m}}$

$K_b$ = Boltzmann’s constant = $1.38066\times 10^{-23} J/K$

T = temperature of the gas

m = mass of the gas particle.

Given, m = $6.65\times 10^{-27} kg$

T = 6.4 K

Substituting all the given values :

$v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}$

$v_{mp}=162.78 m/s$

162.78 m/s is the most probable speed of a helium atom.

4 0

3 years ago

Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

8 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

3 0

3 years ago

A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw

irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0

3 years ago