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3 years ago

Which statement best describes how the temperature of the oceans surface water varies

1 answer:

8 0

If the temperature is high there is less water because it evaporates if it is cloudy it is more because it doesn't evaporate

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A football player runs from his own goal line to the opposing team's goal line returning to the fifty yard line all i 18.0s calc

ANTONII [103]

assuming the pitch is 100yards long, the player runs 100yards to the other goal then a further 50 yards back to the 50-yard line. So he/she runs 150yards in 18s
150/18 = 8.33yards per second average speed.
Initial velocity = 0, average velocity =8.33
Vav = (Vinitial+Vfinal)/2
Vav = 4.16m/s

3 0

3 years ago

Read 2 more answers

An alpha particle(the nucleus of a helium atom) has a mass of 6.64*10^-27kg and a charge of +2e. What are the magnitude and dire

emmasim [6.3K]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Weight = electric force
mg = qE 

6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E
qE =mg,
E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m

8 0

3 years ago

Read 2 more answers

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, w

Viktor [21]

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

$0.9 = P(|\bar{x}-\mu|$

$0.9 = P(\frac{|\bar{x}-\mu|}{\frac{\sigma}{\sqrt{n}}}$

$0.9 = P(|Z|$

For a confidence level of 90% our $Z_{critic}$ is 1.645

Therefore,

$\frac{0.1}{\frac{\sigma}{\sqrt{n}}} = 1.645$

Substituting for $\sigma = 5$ and re-arrange for n, we have that n is equal to

$n=(\frac{1.645\sigma}{0.1})^2$

$n=\frac{(1.645)^2(0.5)^2}{0.1^2}$

$n=67.65$

$n=68$

We need to make 68 readings for have a probability of 90% and our average is within $0.1\°\frac$

3 0

2 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

3 years ago

A water tank, full of water, has internal dimensions of 1 m x 2 m x 5 m. The density of water is 1000 kg/m³. The volume of the w

MatroZZZ [7]

Answer:

volume of water inside the tank is 10 cubic meter

Explanation:

As we know that the volume of total water inside the tank is given as

$V = L \times H \times W$

here we know that

L = length = 1 m

H = height = 2 m

W = width = 5 m

now we have

$V = 1 \times 2 \times 5$

$V = 10 m^3$

So volume of water inside the tank is 10 cubic meter

3 0

3 years ago