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horrorfan [7]
3 years ago
14

Shuffle a deck of cards and turn over the first card. What is the sample space of the experiment? How many outcomes are in the e

vent that the first card is a heart?
Physics
1 answer:
True [87]3 years ago
5 0

Answer:

The sample space contain 52 entities as shown in explanation.

The event that first card is a heart, contains 13 outcomes as shown in the explanation.

Explanation:

The sample space will contain 52 entities as follows:

Sample Space = {Heart 1, Heart 2,Heart 3,Heart 4,Heart 5,Heart 6,Heart 7,Heart 8,Heart 9,Heart Jack,Heart Queen, Heart King,Heart A,Spade 1, Spade 2,Spade 3,Spade 4,Spade 5,Spade 6,Spade 7,Spade 8,Spade 9,Spade Jack,Spade Queen, Spade King,Spade A, Diamond 1, Diamond 2,Diamond 3,Diamond 4,Diamond 5,Diamond 6,Diamond 7,Diamond 8,Diamond 9,Diamond Jack,Diamond Queen, Diamond King,Diamond A, Club 1, Club 2,Club 3,Club 4,Club 5,Club 6,Club 7,Club 8,Club 9,Club Jack,Club Queen, Club King,Club A}

There will be 13 outcomes in the event that first card is a heart, as follows:

Event(first card hearts) = {Heart 1, Heart 2,Heart 3,Heart 4,Heart 5,Heart 6,Heart 7,Heart 8,Heart 9,Heart Jack,Heart Queen, Heart King,Heart A}

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Answer:

Demagnetization processes include heating past the Curie point, applying a strong magnetic field, applying alternating current, or hammering the metal.

Explanation:

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3 years ago
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object
Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

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What does an atomic nucleus give off a particle?
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The correct answer is answer choice B.
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3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

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W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

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3 years ago
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