All categories

3 years ago

PLZ HELP I DONT GET IT

Physics

1 answer:

GaryK [48]3 years ago

6 0

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

You might be interested in

Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high). Express your a

ankoles [38]

Answer:

It will take 8.80 sec to fall from the building

Explanation:

We have given height pf the state building h = 380 m

Initial velocity will be 0 m /sec

So u = 0 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the fall time

According to second equation of motion $h=ut+\frac{1}{2}gt^2$

So $380=0\times t+\frac{1}{2}\times 9.8\times t^2$

$t^2=77.55$

t = 8.80 sec

3 0

3 years ago

An egg of mass 0.060 kg is dropped from the top of a building. Just before it reaches the ground, it has a total kinetic energy

3241004551 [841]

The velocity is 14 m/s

The parameters given on the question are

mass= 0.060 kg

kinetic energy= 5.9 joules

K.E= 1/2mv²

5.9= 1/2 × 0.060 × v²

5.9= 0.5 × 0.060v²

5.9= 003v²

v²= 5.9/0.03

v²= 196.66

v= √196.66

v= 14 m/s

Hence the velocity of the egg before it strikes the ground is 14 m/s

brainly.com/question/2084569?referrer=searchResults

3 0

2 years ago

A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s

tamaranim1 [39]

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

$vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j$

(a) The velocity of boat with respect to ground is

$\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s$