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2 years ago

The density of ice is approximately 0.92 g/cm3 . If a block of ice has a volume of 2300 cm3, what is its mass?

Mathematics

1 answer:

Mumz [18]2 years ago

7 0

Answer:

The mass of the ice block is 2116 grams ⇒ a

Step-by-step explanation:

The formula of density is $p=\frac{m}{V}$ , where m is the mass and V is the volume

To find the mass from this formula multiply both sides by V,

then m = p.V

∵ The density of ice is approximately 0.92 g/cm³

∴ p = 0.92

- That means the mass of 1 cm³ is 0.92 g

∵ The volume of an ice block is 2300 cm³

∴ V = 2300

- Use the formula of the mass

∵ m = p.V

- Substitute the values of p and V in it

∴ m = (0.92)(2300)

∴ m = 2116 grams

The mass of the ice block is 2116 grams

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2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0

8 months ago

11. Which could be the table of values that was used to graph the function of x shown below?

frutty [35]

Answer:

Step-by-step explanation:

Here, we want to select the possible table of the given line

Let us look at the intercepts

On the x-axis, we have an intercept at x = 2

On the y-axis, we have an intercept at y = -4

This means that when y = 0, x = 2 and when x = 0, y = -4

Looking at the options, the table that supports this answer is the table of option D

8 0

2 years ago

A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately <u>923.558 m²</u>

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

2 years ago

-1/2 + 4/9

stira [4]

Answer:

-1/18

Step-by-step explanation:

First, you have to change both fractions to have like denominators. The LCM of 2 and 9 is 18.

Second, you have to multiply -1/2 by 9/9 to get a denominator of 18. -1/2 x 9/9 is - 9/18.

Third, you have to multiply 4/9 by 2/2 to get a denominator of 18. 4/9 x 2/2 = 8/18.

Fourth, -9/18 + 8/18 = -1/18

Hope this helps!

4 0

2 years ago

Answer true or false for 1-10 (picture above)

Softa [21]

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3 years ago