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kipiarov [429]
4 years ago
5

A gas has an initial volume of 212 cm3 at a temperature of 293 K and a pressure of 0.98 atm. What is the final pressure of the g

as if the volume decreases to 196 cm3 and the temperature of the gas increases to 308 K?

Physics
1 answer:
Marizza181 [45]4 years ago
7 0
Using the pressure law (P1 x V1)/ T1 = (P2 x V2)/ T2 where P1= the initial pressure V1= initial volume T1= initial temperature and P2= the final pressure V2= the final volume T2 = the final temperature and temperature is always in kelvin

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How much work is done by you when you push car a distance of 10 m while exerting a horizontal
Igoryamba

W = F x d

F: 450 N

d = 10 m

plug your variables into the equation to get W = 450N x 10m

7 0
3 years ago
Ship A is located 4.2 km north and 2.7 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a vel
kotykmax [81]

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, \vec{u_{A}} = 22 km/h towards South = - 22\hat{j}

Velocity of ship B, \vec{u_{B}} = 39 km/h Towards North east at an angle of 36^{\circ} = \vec{u_{B}} = 39sin36^{\circ} \hat{j}

Now, the velocity of ship A relative to ship B:

\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}

\vec{u_{A}} = - 22\hat{j}

\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

Now,

\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}

4 0
3 years ago
The number of times that a wave vibrates in unit is called
kari74 [83]
I believe it's frequency. 
6 0
3 years ago
A motorist drives north for 35.0 minutes at 85.0 kph. He then stops for 15.0 minutes. The motorist then drives 130.0 km in 2.0 h
o-na [289]

Answer:

A. 216.36 \frac{km}{h}

B. 96.56 \frac{km}{h}

Explanation:

Let s_{1} be the distance in first part.

s_{1} = velocity × time

s_{1} = 85 × \frac{35}{60}

s_{1} = 49.58 km

Let s_{2} be the distance in first part.

s_{2} = 130 km

Average velocity = \frac{Total displacement}{Total time}

When second leg of the trip is

A. Toward north

Average velocity = \frac{[tex] s_{1} + s_{2}}{Total time}  [/tex]

Average velocity = \frac{130+49.58}{0.25+0.58}

Average velocity =216.36 \frac{km}{h}

B. Toward south

Average velocity = \frac{[tex] s_{1} - s_{2}}{Total time}  [/tex]

Average velocity = \frac{130-49.58}{0.25+0.58}

Average velocity =96.56 \frac{km}{h}


3 0
3 years ago
How much work is done to move a 10,000-N car 20 meters?
Helga [31]

f = 10000N
distance moved = 20 m
work done = 10000*20 = 200000
6 0
4 years ago
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