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ArbitrLikvidat [17]
3 years ago
12

The data points you have taken on your lab graphs roughly form a straight line. How do you interpret the slope of this line?

Physics
2 answers:
nekit [7.7K]3 years ago
3 0

Answer: Steeper slopes mean lower speeds.

Explanation:

Ann [662]3 years ago
3 0
If it’s a flat slope then it’s c
But if it’s a wavy slope then it’s a
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A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
Anestetic [448]
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
L/8=1.3m/8=0.1625 m

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
q= -3.75\cdot 10^{-9}C
If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
where k_e=8.99\cdot 10^9Nm^2C^{-2} and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.
5 0
3 years ago
A sprinter generates a constant force of 52 N as he runs 100 m. How much work did he do?
LUCKY_DIMON [66]
You just multiply these two numbers. It's 5200J, or 5.2kJ
3 0
3 years ago
Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state
topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

<u>Given  </u>

<u><em>Process 1 ---> 2 </em></u>

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

Volume of point (1) V1   = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant  

<u>Process 2 ---> 3 </u>

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

<u>Process 3 ---> 1 </u>

The relation of the process PV = constant  

<u>Required  </u>

Sketch the processes on the PV coordinates

The work for each process in kJ  

<u>Solution  </u>

The work is defined by  

W=\int\limits^a_b {x} \, dx

<em>a=V2</em>

<em>b=V1</em>

<em>x=P</em>

<em>dx=dV</em>

<u>Process 1 ---> 2  </u>

P3 = P4 = 5 bar  

W=\int\limits^a_b {x} \, dx

<em>a=V3</em>

<em>b=V2</em>

<em>x=4</em>

<em>dx=dV</em>

putting the value of a, b, x, dx in above integral

W=400 kJ

<u>Process 2 ---> 3 </u>

V = constant Then there is no change in the volume,hence W = 0 kJ  

<u>Process 3 ---> 1  </u>

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1  

 W=\int\limits^a_b {x} \, dx

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

  = -160.944 KJ

5 0
3 years ago
A train moves at 300 km/h. How far will it go in 5 hours?
faust18 [17]

Answer:

1,500 km

Explanation:

300 km/h

300 kilometers PER <u><em>HOUR</em></u>

<u><em></em></u>

So you want to see how far it will go in 5 hours, all you need to do is multiply the kilometers by 5. Since we know that per hour it's 300 km.

300 * 5 = 1,500

The train moves at 1,500 km in 5 hours.

Hope this helped!

Have a supercalifragilisticexpialidocious day!

5 0
3 years ago
Explain partial eclipse and total eclipse for lunar eclipse in detail and summarise
melomori [17]

Answer:

A total eclipse occurs when the dark silhouette of the Moon completely obscures the intensely bright light of the Sun, allowing the much fainter solar aureole to be visible. During any one eclipse, totality occurs at best only in a narrow track on the surface of Earth. This narrow track is called the path of totality.

A partial lunar eclipse happens when part of the Moon enters Earth's shadow. In a partial eclipse, Earth's shadow appears very dark on the side of the Moon facing Earth.

7 0
3 years ago
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