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borishaifa [10]

3 years ago

13

a car travel uniformly speed of 30 km/h for 30 minutes and then at uniform speed of 60km/h for next 30min calculate the average

speed of car

1 answer:

algol133 years ago

3 0

Explanation:

i hope this will help you

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"Which missing item would complete this alpha decay reaction?<br> "

Helga [31]

Find the attachment below.

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3 years ago

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notka56 [123]

The volume of the gas once it reaches the surface of water is 2 liters.

The volume of the air in balloon at depth of 100ft (30m) is 500ml.

The pressure at this point is 4 atm.

Assuming that the balloon have no compression by rubber of balloon the volume of air at the water surface is V.

The pressure at the surface of water is 1 atms.

As we know, from the ideal gas equation,

PV = nRT

Where,

P is the Pressure of gas,

V is the volume of the gas,

n is the number of moles,

R is the gas constant whose value is 0.082057 L atm mol-1 K-1,

T is the temperature.

Assuming that the temperature is constant,

We know,

PV = nRT

All quantities on the right side are constants,

So, we can write,

P₁V₁ = P₂V₂

Putting all the the values,

4(0.5) = 1V₂

V₂ = 2 Liters.

The volume of the air at the surface is 2 liters.

1 liter = 0.001 m

Hence,

2 liters = 0.002 m³

So the volume of air at the surface is 0.002m³.

To know more about Ideal gas Equation, visit,

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3 0

1 year ago

5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.

Assoli18 [71]

Answer:

1kg =1000g

2.5kg

D=12m

t=11s

F=2.5KG

Explanation:

work done =f.d

=2.5×12

=30Nm

55-30

average speed

final - initial

divide by time t(s)

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2 years ago

Which are considered noble gases

sladkih [1.3K]

Answer:Argon

Neon

Helium

Krypton

Xenon

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2 years ago

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A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0

3 years ago