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3 years ago

Which is an example of a mixture? A. water B. trail mix C. hydrogen D. carbon dioxide

Physics

1 answer:

chubhunter [2.5K]3 years ago

6 0

B. trail mix

because you can easily take all the peaces out unlike water you cant physically take the oxygen out

please make me the brainliest

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In a lab, the mass of object a is 52 kg. object a weighs

aksik [14]

If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N

5 0

3 years ago

Read 2 more answers

A person consumes a snack containing 14 food calories (14kcal). what is the power this food produces if it is to be "burned off"

Inessa05 [86]

Answer:

B) 2.7W

Explanation:

Converting Cal to Joule

1 cal = 4.186J

14 kcal = 14 x 1000 x 4.186

= 58604 J

Converting hour to seconds

6 hours = 6 x 60 x 60 seconds

= 21600 seconds

Power is the time rate of doing work.

Power = Work/Time

P = (58604) / (21600)

P = 2.7W

8 0

3 years ago

Read 2 more answers

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago

Why does the evaporation of sweat cool down your skin?

Ne4ueva [31]

So sweat helps cool you down two ways. First, it makes your skin feel cooler when it's wet. And when it evaporates it removes some heat. But sweat will only evaporatoin an environment where there isn't much water in the air.

6 0

3 years ago

Read 2 more answers

The atmosphere of Mars is almost all carbon dioxide and the average surface pressure is 610 Pa (as compared with 101,000 Pa on E

Karolina [17]

Answer:

z = 3,737 10⁵ m

Explanation:

a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation

P V = n R T

P = (n r / V) T

We replace

P = (n R / V) T₀ $e^{- C z}$

b) Let's apply this equation in the points

Lower

.z = 0

P₀ = 610 Pa

P₀ = (nR / V) T₀

Higher.

P = 10 Pa

P = (n R / V) T₀ e^{- C z}

We replace

P = P₀ e^{- C z}

e^{- C z} = P / P₀

C z = ln P₀ / P

z = 1 / C ln P₀ / P

Let's calculate

z = 1 / 1.1 10⁻⁵ ln (610/10)

z = 3,737 10⁵ m

4 0

4 years ago