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Lapatulllka [165]

3 years ago

8

Object B has a mass of 10kg. Object b collides with another object. If the momentum of object b after collision is 18kg•m/s18kg.

m/s, its volocity is_____m/s

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

Step-by-step explanation:

You might be interested in

Element X decays radioactively with a half life of 12 minutes. If there are 160

Salsk061 [2.6K]

Answer:

It would take 75.8 minutes for the element to decay to 2 grams.

Step-by-step explanation:

The number of grams of element x, after t minutes, is given by the following equation:

$X(t) = X(0)(1-r)^{t}$

In which X(0) is the initial amount and r is the decay rate.

There are 160 grams of Element X

This means that X(0) = 160.

So

$X(t) = X(0)(1-r)^{t}$

$X(t) = 160(1-r)^{t}$

Half life of 12 minutes.

This means that X(12) = 0.5*X(0) = 0.5*160 = 80. So

$X(t) = 160(1-r)^{t}$

$80 = 160(1-r)^{12}$

$(1 - r)^{12} = 0.5$

$\sqrt[12]{(1 - r)^{12}} = \sqrt[12]{0.5}$

$1 - r = 0.9438$

So

$X(t) = 160(1-r)^{t}$

$X(t) = 160(0.9438)^{t}$

How long would it take for the element to decay to 2 grams?

This is t for which X(t) = 2. So

$X(t) = 160(0.9438)^{t}$

$2 = 160(0.9438)^{t}$

$(0.9438)^{t} = \frac{2}{160}$

$\log{(0.9438)^{t}} = \log{\frac{2}{160}}$

$t\log{0.9438} = \log{\frac{2}{160}}$

$t = \frac{\log{\frac{2}{160}}}{\log{0.9438}}$

$t = 75.8$

It would take 75.8 minutes for the element to decay to 2 grams.

6 0

3 years ago

What is the sum of 3x10

marusya05 [52]

The answer to 3 x 10 is 30. Think of it as 3 groups of 10. Each of the 3 groups has 10.

8 0

3 years ago

Read 2 more answers

Section 2 Topic 3 Homework

VMariaS [17]

Answer:

65$

Step-by-step explanation:

Section 2 Topic 3 Homework

Question 2 of 4

Miguel spends $100 on books, five of which are hardcover and the remaining four are paperback. The prices at the bookstore are based

on whether the book is hardcover or paperback. It charges $2 less for paperback books than for hardcover books. What is the price for

hardcover books?

4 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.05 in. A one-sample t-test will be carried out

satela [25.4K]

Answer:

a) H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

b) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

c) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

Step-by-step explanation:

Part a

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part b

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part c

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

8 0

3 years ago

Sketch the sinusoidal function that passes through (0,-6) , has a period of 12 and an amplitude of 3

riadik2000 [5.3K]

The sinusoidal function that passes through (0,-6), has a period of 12 and an amplitude of 3 will be y = 3 sin 12x - 6.

What is a sinusoidal Function?

It is a function that repeats itself in a particular time interval.

The equation is given as

y = Asin (ωx + Ф) + k

We have

Amplitude (A) = 3

Period (ω) = 12

Phase difference (Ф) = 0

Then the equation will be

y = 3 sin 12x + k

The function is passing through (0, -6). Then we have

-6 = 3sin (12 × 0) + k

-6 = k

Then the equation will be

y = 3 sin 12x - 6

More about the sine function link is given below.

brainly.com/question/3876065

#SPJ1

4 0

2 years ago