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3 years ago

3. A museum curator moves artifacts into place on various different display

Physics

1 answer:

Alex3 years ago

4 0

Explanation:

a. moving a 145 kg aluminum sculpture across a horizontal steel platform

b. pulling a 15 kg steel sword across a horizontal steel shield

c. pushing a 250 kg wood bed on a horizontal wood floor

d. sliding a 0.55 kg glass amulet on a horizontal glass display case

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What are the 2 types of shadow<br> cast by the moon during a solar<br> eclipse?

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Answer:

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Explanation:

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As a result of air at the top of a barometer the height of the mercury column is 73.5cm when it should be 75.0 cm the volume of

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Explanation:

Using p1p1=p2p2

(75-73.5)x8=(X - 74)X 6

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12=6x-444 (add 12)

6x=456 (divide by 6)

X= 76cm

Therefore the answer for the barometric height is 76cm

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Which part of the eye controls whether or not the pupil dilates or constricts? iris retina optic nerve cornea

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A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerat

Readme [11.4K]

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

$2t^2+92-1200=0$

$t=10\ sec$

We need to calculate the velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=92+4\times10$

$v=132\ m/s$

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.

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4 years ago

9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

$V_1$ = Volumen inicial = 95 mL

$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}$

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0

3 years ago