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givi [52]
2 years ago
8

An image of the moon is focused onto a screen using a converging lens of focal length f= 34.3 cm. The diameter of the moon is 3.

48×10^6 m, and its mean distance from the earth is 3.85×10^8 m. What is the diameter of the moon's image?
Physics
1 answer:
Rashid [163]2 years ago
7 0

Answer:

The diameter of the moon's image is 0.31 cm.

Explanation:

Given that,

Focal length = 34.3 cm

Diameter of the moon d=3.48\times10^{6}\ m

Mean distance from the earth d=3.85\times10^{8}\ m

At that distance the object is assumes to be at infinity. hence the image will be formed at a distance equal to focal length

So, the image distance is 34.3 cm.

We need to calculate the  diameter of the moon's image

Using formula of magnification

m= \dfrac{image\ distance}{object\ distance}=\dfrac{height\ of\ image}{height\ of\ object}

\dfrac{0.343}{3.85\times10^{8}}=\dfrac{h'}{3.48\times10^{6}}

h'=\dfrac{0.343\times3.48\times10^{6}}{3.85\times10^{8}}

h'=0.0031\ m= 0.31\ cm

Hence, The diameter of the moon's image is 0.31 cm.

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Answer:

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Explanation:

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vy = 75 m/s

(a) Angular momentum is given by

\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

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\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}

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(b) Here, \overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}

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Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2

Moment of inertia is given by

M=\frac{1}{2}mr^2\\\Rightarrow M=\frac{1}{2}\times 0.8\times 0.135^2\\\Rightarrow M=0.00729\ kgm^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=0.00729\times -0.51312\\\Rightarrow \tau=-0.0037406448\ Nm

The torque the friction exerts is -0.0037406448 Nm

For more information on torque and moment of inertia refer

brainly.com/question/13936874

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If you increase the mass but keep force constant what will happen to the acceleration of the object
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Ft = 758N

Work done Wd = Ft * distance

Wd = 785 * 4 = 3.14 * 10^3J

3 0
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