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Leya [2.2K]
2 years ago
8

An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 a

nd the temperature remains constant?
Physics
1 answer:
sveticcg [70]2 years ago
5 0

Answer:

<h2>2.22 kPa</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{0.4 \times 500000}{0.9}   =  \frac{200000}{0.9} \\  = 222222.2222... \\  = 222222

We have the final answer as

<h3>2.22 kPa</h3>

Hope this helps you

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in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?​
Pavlova-9 [17]

Answer:

There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.

Explanation:

3 0
3 years ago
A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a
Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}

From Newton's second law

F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}

Squarring both sides

112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m

The height from which the student fell is 5.7141 m

5 0
3 years ago
An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
A ball is dropped from rest. how fast is the ball going after 3 seconds​
Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

5 0
3 years ago
How much time is needed to produce 720 Joules of work if 90 watts of power is used?
Tems11 [23]

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

Power=\frac{work.done}{time.taken} \\ \\ time.taken=\frac{work.done}{power}\\=\frac{720J }{90W} \\ \\ =8 s

3 0
3 years ago
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