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Katen [24]
3 years ago
14

How would I label the y axis and fill it out and answer questions at the bottom I need answers please and thanks

Physics
1 answer:
yKpoI14uk [10]3 years ago
7 0

you must write the temprature of the copper sulfate solution on the on the y axis

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0
3 years ago
A 64.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxyg
Umnica [9.8K]

dfdfdfdddfdddddddddddddddd

6 0
3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

Distance covered in the slowing down phase = 250 m

v^2-u^2=2as

a= \frac {(v^2-u^2)}{2s}

a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2

v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

The car is in the pit stop for 5s t_2=5 s

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

v=u+at

t= \frac{(v-u)}{a}

t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0
3 years ago
A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 3.84 m/s
JulijaS [17]

Answer:

the train is moving at the speed of v = 1.79 m/s

Explanation:

given,

rain drop is falling vertically down with the speed of  = 3.84 m/s

angle of the rain drop = 25°                      

tan θ = \dfrac{v}{u}                      

tan 25° = \dfrac{v}{3.84}                      

v =3.84 × tan 25°                      

v = 1.79 m/s                  

hence, the train is moving at the speed of v = 1.79 m/s

3 0
3 years ago
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