All categories

2 years ago

A ball bearing has mass 0.44 pounds. 1 kg = 2.2 pounds.

Mathematics

1 answer:

torisob [31]2 years ago

7 0

Answer:

(i) 0.2 kg

(ii) kg/cm³

(b) 2.29 kg

Step-by-step explanation:

This problem gives you relations between units, and the formulas for density and mass. It is asking you to make use of these relations in various mass and density calculations.

The relationship between kg and pounds can be used to write a units conversion multiplier. It will have numerator equal to denominator, but will have units such that multiplying by it will give the units you want.

0.44 pounds × multiplier = __ kg

0.44 pounds × (1 kg)/(2.2 pounds) = (0.44/2.2)(1 kg)

0.44 pounds = 0.2 kg

The formula can be used to find the units of interest:

Density = mass/volume

units of density = (units of mass)/(units of volume)

units of density = kg/cm³

Using the given formula with the given value of L, we find the volume to be ...

V = 4(1.40 cm)(3 -1.40 cm)² = 14.336 cm³

Then the mass is ...

mass = volume × density

mass = (14.336 cm³)×(0.160 kg)/(1 cm³) = 2.29376 kg

The mass is about 2.29 kg.

_____

<em>Additional comment</em>

Often you will see density in terms of kg/m³ or g/cm³. Water has a density of about 1 g/cm³ or 1000 kg/m³. The most dense non-radioactive element in the periodic table is osmium, with a density of about 22.6 g/cm³, or 0.0226 kg/cm³. The material in part (b) of this problem has a density about 7 times that, so is clearly not from Earth.

You might be interested in

Plllzz hellpppp meeeee thhhhxx

Naddik [55]

Most likely it’s x=7 unless there are other options you didn’t show

6 0

3 years ago

Read 2 more answers

For each of the following, use logic laws to decide whether the statement is a tautology contradiction, or neither. a (a) (A B)

spayn [35]

Answer:

The statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

The statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

Step-by-step explanation:

A tautology is a proposition that is always true.

A contradiction is a proposition that is always false.

A contingency is a proposition that is neither a tautology nor a contradiction.

a) To classify the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ , you need to use the logic laws as follows:

$(A\rightarrow \lnot B)\land (B\rightarrow A) \equiv$

$\equiv (\lnot A \lor\lnot B)\land(\lnot B \lor A)$ by the logical equivalence involving conditional statement.

$\equiv (\lnot B\lor \lnot A )\land(\lnot B \lor A)$ by the Commutative law.

$\equiv \lnot B \lor (\lnot A \land A)$ by Distributive law.

$\equiv \lnot B \lor (A \land \lnot A)$ by the Commutative law.

$\equiv \lnot B \lor F$ by the Negation law.

Therefore the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

b) To classify the statement $(P\rightarrow \lnot P)\land P$ , you need to use the logic laws as follows:

$(P\rightarrow \lnot P)\land P \equiv$

$\equiv (\lnot P \lor \lnot P)\land P$ by the logical equivalence involving conditional statement.

$\equiv P \land (\lnot P \lor \lnot P)$ by the Commutative law.

$\equiv (P \land \lnot P) \lor (P \land \lnot P)$ by Distributive law.

$\equiv F \lor F \equiv F$ by the Negation law.

Therefore the statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

8 0

3 years ago

Given a quadratic function in vertex form:

Alja [10]

Answer:

See below

Step-by-step explanation:

A vertical stretch takes place when $a\cdot f(x)$ given that $|a| > 1$

A vertical compression takes place when $a\cdot f(x)$ given that $0 < |a| < 1$

A vertical stretch is different than a horizontal compression:

In a vertical stretch, the input stays the same, but the output is multiplied by the scale factor
In a horizontal compression, the output stays the same, but the input is multiplied by the scale factor

A reflection across the x-axis means that the output, our y-variable, is the opposite sign. This means that all values of $a$ must be negative such that $a\cdot f(x)$ as mentioned in parts A and B. Also, in part C, since our scale factor is negative, the output is the only one being multiplied by the scale factor.

7 0

2 years ago

An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty y

Kay [80]

Answer:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}$

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

Step-by-step explanation:

Information given

$\bar X_{1}=14$ represent the mean for sample 1 (younger)

$\bar X_{2}=20$ represent the mean for sample 2 (older)

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=31$ sample size for the group 2

$n_{2}=31$ sample size for the group 2

t would represent the statistic

System of hypothesis

We want to test if that people under the age of forty have vocabularies that are different than those of people over sixty years of age, the system of hypothesis are:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

The statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=31+31-2=60$

Replacing the info given we got:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}$

6 0

3 years ago

Select all the equations on which the point (10,0) lies.

Makovka662 [10]

Answer:

B, C

Step-by-step explanation:

Since y=0, you are looking for equations such that 10 times the coefficient of x is the constant in the equation.

a) 5·10 ≠ 15

b) 2·10 = 20 . . . . select this one

c) 10 = 10 . . . . select this one

d) 3·10 ≠ 13

e) 4·10 ≠ 20

f) 6·10 ≠ 50

8 0

2 years ago

A ball bearing has mass 0.44 pounds. 1 kg = 2.2 pounds. ​

A ball bearing has mass 0.44 pounds. 1 kg = 2.2 pounds.