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2 years ago

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1.Narysuj obraz ołówka o długości 6 cm znajdującego się w odległości 8 cm od soczewki o ogniskowej 6 cm

1 answer:

Aliun [14]2 years ago

5 0

Answer:

free quotes from carpet repair contractors at door in

Explanation:

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If a battery is running down, a light bulb in the circuit will become dimmer because when voltage drops

lana [24]

A. There is less current flowing through the bulb.

Ohms law dictates that Electrical voltage = current x resistance. To understand current you would divide voltage by resistance. The less voltage the less current will run through the circuit.

Mark as brainliest if satisfied.

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3 years ago

The semi-major axis of this ellipse is 8.8 cm, and the distance from one of the foci to the

scoundrel [369]

The eccentricity is 0.5

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3 years ago

A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an

BARSIC [14]

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

Therefore, rhe normal force is equivalent to 1,373.4 N

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3 years ago

If the mass is 22 g and the volume is 11 ml, what is the density of the object

andriy [413]

2 g/mL there’s your answer

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3 years ago

Read 2 more answers

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

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3 years ago