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3 years ago

Please help i'm going to throw up from stress

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

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You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

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3 years ago

Which of the following statements are true?Check all that apply.

MaRussiya [10]

Answer:

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

Explanation:

As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path

So all planets are in elliptical path always

Position of sun is always at one of the focus

so correct answer will be

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

4 0

3 years ago

Physics. Need help. Brainlieast answer for most/ all of the answers answered

Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s; b). 29.7 m/s

10). 7.24 meters

11). 700 meters

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3 0

3 years ago

A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant

lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

$N = 1201$

$SS = 10800$

Required

Determine the standard deviation

First, we need to determine the variance;

$Variance = \frac{SS}{N - 1}$

This gives:

$Variance = \frac{10800}{1201 - 1}$

$Variance = \frac{10800}{1200}$

$Variance = 9$

Know that:

$Variance = SD^2$

Where SD represents standard deviation

This gives

$9 = SD^2$

Take square root

$SD = \sqrt 9$

$SD = 3$

7 0

3 years ago

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

3 years ago