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The isotope cobalt-60 has a nuclear mass of 59.933820 u

Anit [1.1K]

To start this question, we should know what is the atomic number of cobalt. The atomic number (the number of protons) of Cobalt is Z =27.

Now, we know that a Cobalt 60 isotope means an isotope of Cobalt whose Atomic Mass is 60.

Thus, in a Cobalt 60 isotope, the number of neutrons in the nucleus are

$60-Z=60-27=33$

From the question we know that the given nuclear mass is 59.933820 u.

Now, the mass defect of Cobalt 60 can be easily calculated by adding the masses of the protons and the neutrons as per our calculations and subtracting the given nuclear mass from it.

Thus,

Mass Defect = (Number of Protons Mass of Proton given in the question) + (Number of Neutrons Mass of Neutron given in the question)-59.933820 u

$\therefore 27\times 1.007825+33\times 1.008665-59.933820=0.5634 u$

Thus, the required Mass Defect is 0.5634u.

In eV, the Mass Defect is $931.5MeV\times 0.5634=534.807 MeV$

6 0

3 years ago

Read 2 more answers

14. The diagonals of square ABCD intersect at E. If AE = 2, find the perimeter of ABCD.

lutik1710 [3]

Answer:

B, $8\sqrt{2}$

Step-by-step explanation:

Diagonals of a square are congruent so AC= 4

using pythagorean theorem we can then do $x^{2} +x^{2}$ = $4^{2}$

$2x^{2} = 16\\x^{2} =8\\x=\sqrt{8} \\x=2\sqrt{2}$

Then for perimeter we times the $2\sqrt{2}$ by 4 and get

$8\sqrt{2}$

Hope this helps and please mark brainliest!!

8 0

3 years ago

A Statistics professor has observed that for several years students score an average of 106 points out of 150 on the semester ex

tekilochka [14]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 106

For the alternative hypothesis,

µ > 106

It is a right tailed test because of >.

Since no population standard deviation is given, the distribution is a student's t.

Since n = 241,

Degrees of freedom, df = n - 1 = 241 - 1 = 240

t = (x - µ)/(s/√n)

Where

x = sample mean = 109

µ = population mean = 106

s = samples standard deviation = 8.3

a) test statistic, t = (109 - 106)/(8.3/√241) = 5.61

We would determine the p value using the t test calculator. It becomes

p = 0.00001

Since alpha, 0.05 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 5% level of significance, this improvement seem to be practicallysignificant.

b) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

n = number of samples

From the information given, the population standard deviation is unknown, hence, we would use the t distribution to find the z score

df = 240

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 1.6512

Margin of error = 1.6512 × 8.3/√241

= 0.88

Confidence interval = 109 ± 0.88

7 0

3 years ago

Which of the following vectors are orthogonal to (1,5)? Check all that apply

Andre45 [30]

We know that
If the scalar product of two vectors is zero, both vectors are orthogonal

A. (-2,5)
(-2,5)*(1,5)-------> -2*1+5*5=23-----------> are not orthogonal

B. (10,-2)
(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are orthogonal

C. (-1,-5)
(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal

D. (-5,1)
(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are orthogonal

the answer is
B. (10,-2) and D. (-5,1) are orthogonal to (1,5)

7 0

3 years ago

Explain how you know 21/100 is greater than 1/5

Nataly_w [17]

This problem can be solved using equivalent fractions. The first step in resolving this problem is to realize that fractions are best compared when they both have the same denominator. In this case, I will choose to make that common denominator 100. There is no need to rewrite the fraction $\frac{21}{100}$ as the denominator for this fraction is already 100. The fraction $\frac{1}{5} =\frac{20}{100}$ . This is achieved by multiplying both the numerator and denominator by 20. Now that the two fractions have the same denominator, we can easily see that $\frac{21}{100}$ is greater than $\frac{1}{5}$ because it is greater than its equivalent fraction.

5 0

3 years ago