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Licemer1 [7]
4 years ago
10

2. The following formula relates three quantities: Force (F), mass (m), and acceleration (a).

Mathematics
2 answers:
castortr0y [4]4 years ago
5 0

Let

F--------> represent the force

m--------> represent the mass

a-------> represent the acceleration

we know that

The formula that relates three quantities: Force (F), mass (m), and acceleration (a) is equal to

F=ma

<u>Part a)</u> Solve this equation for a

F=ma

Divide both sides by the mass

a=\frac{F}{m}

therefore

<u>the answer part a) is equal to</u>

a=\frac{F}{m}

<u>Part b) </u>Let the force be 25 units and the mass be m = 10 units. What is the acceleration, a?

we know that

the formula of acceleration is equal to

a=\frac{F}{m}

in this problem

we have

F=25\\m=10

Substitute the values n the formula

a=\frac{25}{10}

a=2.5

therefore

<u>the answer part b( is equal to</u>

a=2.5

<u>Part c)</u> Let the force be F = 25 units and the acceleration be a = 5 units. What is the mass, m?

we know that

F=ma

Solve for m

Divide both sides by the acceleration

m=\frac{F}{a}

in this problem

we have

F=25\\a=5

Substitute the values n the formula

m=\frac{25}{5}

m=5

therefore

<u>the answer Part c) is </u>

m=5

Ierofanga [76]4 years ago
3 0
(A): Divide each side by m and you will get \frac{F}{m} = a

(B): (You didn't tell me the units of force, so I can't give you the answer)

(C): m = 
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If you are dealt five cards from a shuffled deck of 52 cards find the probability of getting three queens and two kings
dexar [7]

The probability of getting three queens and two kings is \frac{1}{1082900}

<u>Solution:</u>

Given that , you are dealt five cards from a shuffled deck of 52 cards  

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Now, we know that, in a deck of 52 cards, we will have 4 queens and 4 kings.

\text { probability of an event }=\frac{\text { favarable possibilities }}{\text { number of possibilities }}

<em><u>Probability of first queen:</u></em>

\text { Probability for } 1^{\text {st }} \text { queen }=\frac{4}{52}=\frac{1}{13}

<em><u>Probability of second queen:</u></em>

\text { Plobability for } 2^{\text {nd }} \text { queen }=\frac{3}{51}=\frac{1}{17}

Here we used 3 for favourable outcome, since we already drew 1 queen out of 4

And now number of outcomes = 52 – 1 = 51

<em><u>Probability of third queen:</u></em>

Similarly here favorable outcome = 2, since we already drew 2 queen out of 4

And now number of outcomes = 51 – 1 = 50

\text { Probability of } 3^{\text {rd }} \text { queen }=\frac{2}{50}=\frac{1}{25}

<em><u>Probability for first king:</u></em>

Here kings are 4, but overall cards are 49 as 3 queens are drawn

\text { probability for } 1^{\text {st }} \text { king }=\frac{4}{49}

<em><u>Probability for second king:</u></em>

Here, kings are 3 and overall cards are 48 as 3 queens and 1 king are drawn

\text { probability of } 2^{\text {nd }} \text { king }=\frac{3}{48}=\frac{1}{16}

<em><u>And, finally the overall probability to get 3 queens and 2 kings is:</u></em>

=\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{4}{49} \times \frac{1}{16}=\frac{4}{4331600}=\frac{1}{1082900}

Hence, the probability is \frac{1}{1082900}

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