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kondaur [170]
4 years ago
13

Thomas wants to use his 5.5 D reading glasses as a simple magnifier. What is the angular magnification of this lens when Thomas'

eye is relaxed?
Physics
1 answer:
juin [17]4 years ago
3 0
Sorry, but the given info is insufficient. 
<span>The angular magnification is the ratio of the viewing distances. </span>
<span>We assume the greater viewing distance is 26 cm. </span>
<span>But the nearer viewing distance depends on the focal length of Thomas's <relaxed> eye in combination with the glasses. We don't know about his relaxed eye. We'd need to know his far point. </span>
<span>For instance, say his far point is 2.5 m. Then his focal length with glasses = 1/(1/2.5+6.2) m = 15.1515 cm. Magnification = 26/15.1515 = 1.716. </span>
<span>If his far point is at infinity, magnification = 1.612.</span>
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1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
balu736 [363]

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

5 0
3 years ago
If a honeybee is flying 7.0 m/s, what is the kinetic energy (Joules) if its mass is 0.1 g?
svetlana [45]

Answer:

KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules

Explanation:

mass=m=0.1g=0.0001kg

velocity=v=7.0m/s

kinetic energy=K.E.=?

as we know that

kinetic energy=\frac{1}{2}mv^2\\ putting values\\kinetic energy=\frac{1}{2}(0.0001kg)(7.0m/s)^2\\ kinetic energy=\frac{1}{2} 0.0049joules\\kinetic energy= 0.00245Newton meter or joules

hope it will help ^_^

5 0
3 years ago
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0
3 years ago
A slit 0.240mm wide is illuminated by parallel light rays with a wavelength 600nm . The diffraction pattern is observed on a scr
taurus [48]

To solve this exercise, it is necessary to apply the concepts on the principle of superposition, specifically on constructive interference,

Constructive interference (light spot) is defined by

dsin\theta=m\lambda

Where,

m = The integer m is called the interference order and is the number of wavelengths by which the two paths differ.

\lambda = wavelenght

d = Distance

For smaller angles sin\theta = tan\theta,

d tan\theta = m \lambda

From the trigonometric properties it is understood that so the is - in this case - the length measured vertically by reason of distance, that is

d*\frac{y}{L} = m\lambda

Re-arrange to find y,

y = \frac{m\lambda L}{d}

Replacing our values

y  = \frac{(1)(600*10^{-9}(3.5m)}{0.240*10^{-3}m}

y = 8.75*10^{-3}m

y = 8.75mm

PART B) To calculate the intensity it is necessary to find the angle between the previously calculated height and distance in order to calculate the phase angle, in other words,

tan\theta' = \frac{y}{L}

tan\theta' = \frac{1/2(8.75*10^{-3})}{3.5}

\tetha = 0.07161\°

Therefore phase angle is

\gamma = \frac{2\pi}{\lambda}sin\theta'

\gamma = \pi

The intensity formula would then be given by,

I = I_0 \frac{sin\frac{\gamma}{2}^2}{\frac{\gamma}{2}}

I = 4.1*10^{-6}(\frac{4}{\pi})

I = 1.66*10^{-6}W/m^2

3 0
4 years ago
Pls help it’s due today
AveGali [126]

Explanation:

so I think it B hope that I helped you can good lick

4 0
3 years ago
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