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3 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?

1 answer:

Marizza181 [45]3 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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A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?

andriy [413]

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

$v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph$

So, the runner's average speed is 22.54 mph.

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3 years ago

What happens to myosin and actin as sarcomeres relax?

Leokris [45]

The myosin heads pull on the actin, bringing them closer together

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3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

What is parkour and how do you do it?

saw5 [17]

Answer:

Parkour is a type of sport or “Hobby”

Explanation:

Parkour is the activity or sport of running through an area, typically in an urban environment.

You can perform parkour by jumping over certain objects, leaping and turning, etc.

Hope this helped!

~Oreo!

4 0

3 years ago

A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the

meriva

Answer:

a. $2.86 m/s^2$

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

$\mu_s=0.14$

$g=9.8 m/s^2$

By Newton's second law

$8F-f=(8M+m)a$

$a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}$

$a=2.86 m/s^2$

b.By Newton's second law

$T=ma+\mu_s mg$

Substitute the values

$T=250\times 2.86+0.14(250)(9.8)=1058 N$

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0

3 years ago