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frez [133]
2 years ago
6

Part 1 out of 2

Mathematics
1 answer:
Hoochie [10]2 years ago
6 0

Answer:

Step-by-step explanation:

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John cut 6 yards of fabric into pieces of equal length. Each piece is the fraction one third yard long. Into how many pieces did
deff fn [24]

Answer:

18 pieces

Step-by-step explanation:

Take the amount of fabric and divide by the length of the piece to determine how many pieces

6 ÷ 1/3

Copy dot flip

6 * 3/1

18 pieces

7 0
2 years ago
Read 2 more answers
Calculate the momentum of a proton moving with a speed of (a) 0.010c, (b) 0.50c, (c) 0.90c. (d) convert the answers of (a)â(c) t
ANTONII [103]

The momentum of a proton would be:

(a) For particle moving with speed 0.010c

p = 5.016 × 10^(-21) kgms^{-1}

p = 9.398 MeV/c

(b) For particle moving with speed 0.50c

p = 2.89 × 10^(-19) kgms^{-1}

p = 541.5 MeV/c

(c) For particle moving with speed 0.90c

p = 23.73 × 10^(-19) kgms^{-1}

p = 4446.35 MeV/c

For given question,

We need to calculate the momentum of a proton moving with a given speed.

We know that, the equation for relativistic momentum is,

p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2} } }

We know, the mass of proton (m) = 1.67\times 10^{-27} kg

(a) For particle moving with speed 0.010c

v = 0.010c

the momentum of a proton would be,

\Rightarrow p=\frac{(1.67\times 10^{-27}kg)\times (0.010\times 3\times 10^8)\frac{m}{s} }{\sqrt{1-\frac{0.01^2~c^2}{1^2~c^2} } }\\\\\Rightarrow p=5.016\times 10^{-21}~~kgms^{-1}

(b) For particle moving with speed 0.50c

v = 0.50c

the momentum of a proton would be,

\Rightarrow p=\frac{(1.67\times 10^{-27}kg)\times (0.50\times 3\times 10^8)\frac{m}{s} }{\sqrt{1-\frac{0.5^2~c^2}{1^2~c^2} } }\\\\\Rightarrow p=2.89\times 10^{-19}~~kgms^{-1}

(c) For particle moving with speed 0.90c

v = 0.90c

the momentum of a proton would be,

\Rightarrow p=\frac{(1.67\times 10^{-27}kg)\times (0.90\times 3\times 10^8)\frac{m}{s} }{\sqrt{1-\frac{0.90^2~c^2}{1^2~c^2} } }\\\\\Rightarrow p=23.73\times 10^{-19}~~kgms^{-1}

Now, we need to convert answers into MeV/c

1 MeV = 1.6 × 10^(-13) kg.m²/s²

⇒ 1  kg.m²/s² = 625 × 10^(10) MeV

1 c = 299,792,458 m/s

⇒ 1 m/s = 3.3356E-9 c

So, 1  kg. m/s = 1.8737259e+21 MeV/c

(a) For p = 5.016 × 10^(-21) kgms^{-1}

⇒ p = 5.016 × 10^(-21) × 1.8737259e+21

⇒ p = 9.398 MeV/c

(b) For p = 2.89 × 10^(-19) kgms^{-1}

⇒ p = 2.89 × 10^(-19) × 1.8737259e+21

⇒ p = 541.5 MeV/c

(c) For p = 23.73 × 10^(-19) kgms^{-1}

⇒ p = 23.73 × 10^(-19) × 1.8737259e+21

⇒ p = 4446.35 MeV/c

Therefore, the momentum of a proton would be:

(a) For particle moving with speed 0.010c

p = 5.016 × 10^(-21) kgms^{-1}

p = 9.398 MeV/c

(b) For particle moving with speed 0.50c

p = 2.89 × 10^(-19) kgms^{-1}

p = 541.5 MeV/c

(c) For particle moving with speed 0.90c

p = 23.73 × 10^(-19) kgms^{-1}

p = 4446.35 MeV/c

Learn more about the momentum here:

brainly.com/question/12942930

#SPJ4

5 0
1 year ago
Please help me with this dont guess pls!
dexar [7]

Answer:

B

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Help me do the ones marked in red.
Dmitrij [34]

Answer:

  2. "two thirds of the fourth power of <em>r</em>"

  4.  n + 14

  6.  7 +11n

  8.  (2/5)n^2

__

  7.  28

  8.  1/5

Step-by-step explanation:

2. "two thirds of the fourth power of <em>r</em>"

__

4. "sum" means the listed items are added. You can use any convenient variable to represent "a number." I often use <em>n</em> for <em>n</em><em>umber</em>.

  n + 14

__

6. "7 more than" means 7 is added to whatever follows. "11 times a number" means the number is multiplied by 11.

  7 + 11n

__

8. "of" means "times". The square of a number is that number to the power of 2.

  (2/5)n^2

__

7 & 8 (evaluation). Your calculator can do this. (If not, get one that can.) Note that parentheses are needed around numerators and denominators and anything else that must be treated as a single value.

8 0
2 years ago
n a call center the number of received calls in a day can be modeledby a Poisson random variable. We know that, on average, 0.5%
guapka [62]

Answer:Pois(ln(200))

Step-byy-step explanation:

Let N be the number of received calls in a day

That is

N∼Pois(λ).

0.5% = 0.5/100 = 1/200 of no calls

P(N=0)=e^−λ=1/200

-λ=e^(1/200)

λ=in(200)

Our number of calls in a day is distributed Pois(ln(200)).

7 0
3 years ago
Read 2 more answers
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