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3 years ago

Hi:) anyone able to help with part - aii the one on cooling fins ? Thankss!

Physics

1 answer:

taurus [48]3 years ago

5 0

The cooling find are there to keep the temperature low. They are mostly black in color, and hence, radiate heat, which is how the temperature is controlled.

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An object is sitting on the floor. A 22.4 N force is pulling the object to the right and an 11 N force is pulling the object to

Nimfa-mama [501]

answer = 33.4 net force.

5 0

2 years ago

Cold plasma used in TVs is different from the solar plasma because ___________.

Arisa [49]

Answer: B i believe

Explanation:

5 0

3 years ago

A lens is used to make an image of magnification -1.50. The image is 30.0 cm from the lens. What is the object’s distance (unit=

jeka57 [31]

Answer: 20cm

Explanation:

Given the following data:

Magnification of image = - 1.50

Distance of image from the lens = 30.0cm

The Magnification of an image is the ratio of the image distance from the lens and the object distance from the lens.

Mathematically,

Magnification(M) = - (image distance (di) / object distance (do))

M = - (di/do)

-1.50 = - (30/do)

do = 30 / 1.5

do = 20 cm

Object is placed at a distance of 20cm from the lens.

5 0

2 years ago

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v

vladimir1956 [14]

Answer:

$P(X\ge 1) = 0.9502$

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$\lambda = density$

$\lambda = \frac{3}{10}$

$\lambda = 0.3$

T = the light years

$T = 10$

Calculating $P(X \ge 1)$

In probability:

$P(X \ge 1) = 1 - P(X = 0)$

Calculating P(X=0)

Substitute 0 for k and the values for $\lambda$ and T in

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}$

$P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}$

$P(X=0) = (3)^0 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-3}$

$P(X=0) = e^{-3}$

$P(X=0) = 0.04979$

Substitute 0.04979 for P(X=0) in $P(X \ge 1) = 1 - P(X = 0)$

$P(X\ge 1) = 1 - 0.04979$

$P(X\ge 1) = 0.95021$

$P(X\ge 1) = 0.9502$ --- approximated

<em>Hence, the required probability is 0.9502</em>

3 0

2 years ago

HELP ME PLS THIS IS DUE TODAY I NEED HELP!!!!

lions [1.4K]

Explanation:

i wish I could help you but its not clear...

4 0

2 years ago