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2 years ago

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A building is 6m high, and it is 80m from a converging camera lens. If the camera forms an image which is 6 mm high, (a) What is

the magnification? (b) How far must the camera film be behind the lens for the image to be formed?

1 answer:

Lostsunrise [7]2 years ago

5 0

Answer:

Explanation:

Two aircrft pand qare flyingatthe same speed 300m the diraction along tothe diraction of the veloctiy

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I need help on 4 and 5 please

Ivan

Answer:

c

Explanation:

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3 years ago

A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t

Kazeer [188]

Answer:

Explanation:

Given

Length of rope $L=55\ m$

Weight of rope $W=4040\ N$

weight density $\lambda =\frac{4040}{55}=73.45\ N/m$

Work done to lift rope 33 m

$W=\int_{0}^{33}\lambda hdh$

$W=\int_{0}^{33}73.45hdh$

$W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0$

$W=39.993\ kJ$

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3 years ago

Microwave transmission involves sending signals from one microwave station to another. What is this often called?

astraxan [27]

It is called fixed wireless

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2 years ago

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago