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xz_007 [3.2K]
3 years ago
14

A straight bar magnet is initially 4 cm long, with the north pole on the right and the south pole on the left. if you cut the ma

gnet in half, the right half will
Physics
1 answer:
Greeley [361]3 years ago
6 0
The right half will be a new bar magnet of 2cm with north pole on the right side and south pole on teh left.
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What is the rabbit's displacement from t = 0s to 3 s?<br> Answer with two significant digits.
Inessa05 [86]

Answer: i think the answer is 20.0s

Explanation:

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Solve this for me please
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Power output = V*I=11000*750=8250 kVA= 8250 kW
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In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit
arlik [135]

Answer:

Equilibrium quantity = 5

Equilibrium price = 40

Explanation:

given:

p = -x²-3x+80

p = 7x+5

For the equilibrium quantity the price from both the functions will be equal

thus, we have

-x² - 3x + 80 = 7x+5

⇒ x² +3x + 7x + 5 - 80 = 0

⇒x² + 10x - 75 = 0

now solving for x

x²- 5x + 15x -75 = 0

x(x-5) + 15(x-5) = 0

therefore, the two roots of the equation are

x = 5 and x = -15

since the quantity cannot be in negative

therefore, the equilibrium quantity will be = 5

now the equilibrium price can be found out by substituting the equilibrium quantity in any of the equation

thus,

p = -(5)² -3(5) + 80 = 40

or

p = 7(5) + 5 = 40

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An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

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

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

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