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3 years ago

HELP!!! I will 5 star if its helpful :(

Physics

1 answer:

victus00 [196]3 years ago

5 0

Answer:

Net Force Formula: F1+F2+F3...+FN

*(MAKE SURE YOU MAKE NOTE OF THE NEGATIVE FORCES AND SUBTRACT THEM!)*

Explanation:

Top left: 17N+25N+25N-42N= 25N

Top right: 65N+200N-65N-150N-200N= 150N

Bottom left: 189N+123N+284N-96N-188N-312N= 0N

Bottom right: 34N+77N-12N-34N= 65N

I hope this helps! :))

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Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.

Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}$

$\dfrac{1}{v}=\dfrac{3}{25}$

$v=8.33\ cm$

We need to calculate the image distance from right lens

The object distance will be

$u = 15-8.33 = 6.67\ cm$

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}$

$\dfrac{1}{v}=\dfrac{1167}{3335}$

$v=2.86\ cm$

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0

3 years ago

What is the equivalent of 0° C in Kelvin?<br> Help will give brainiest

givi [52]

Answer:

273.15 dat is the answer im pretty sure

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3 years ago

(a) Find the acceleration of B.<br>(b) Find the tensions, T1 and T2, in the strings.

Ann [662]

i believ that the answer would be

the acceleration of B is 0.2

6 0

3 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

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3 years ago

Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done

sergiy2304 [10]

A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J

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2 years ago