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bezimeni [28]
3 years ago
13

1. Which letter represents the location of the battery in this diagram?

Physics
2 answers:
dem82 [27]3 years ago
5 0

Answer:

1. The battery is represented by the symbol next to letter A.  The plus and the minus signs indicate the positive and negative poles of the battery, respectively.

2. The switch is represented by the symbol next to letter B. Here, the switch is closing the circuit, that is, there is electric current circulating.

My name is Ann [436]3 years ago
4 0

Answer:

location of battery in this diagram is at A and location of switch is at B.

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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp
Bond [772]

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force  

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

7 0
3 years ago
Consider a spherical capacitor with radius of the inner conducting sphere a and the outer shell b. The outer shell is grounded (
AleksAgata [21]

Answer:

Explanation:

The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.

The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)

6 0
3 years ago
Bádminton is played to a score of
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Badminton is played to a score of 21 points
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What is carbon dating
Tom [10]

the determination of the age or date of organic matter from the relative proportions of the carbon isotopes carbon-12 and carbon-14 that it contains. The ratio between them changes as radioactive carbon-14 decays and is not replaced by exchange with the atmosphere.

7 0
3 years ago
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as
Leona [35]

(a) -48.0 cm, diverging

We can use the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}

f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm

The lens is diverging, since the focal length is negative.

(b) 6.38 mm, erect

We can use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the size of the image

y = 8.50 mm is the size of the object

Substituting p and q that we used in the previous part of the problem, we find y':

y'=-y\frac{q}{p}=-(8.50 mm)\frac{-12.0 cm}{16.0 cm}=6.38 mm

and the image is erect, since the sign is positive.

(c)

See attached picture.

4 0
3 years ago
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