What is a homogeneous mixture in which particles never separate?

Can someone please tell me...... What does it mean if you through up thew your nose and mouth at the same time!?

hodyreva [135]

if you're worried about youre health you're completely fine. (other than the fact you are throwing up, if your sick please see a doctor) but if the things you're worried about is that it can out of your nose, dont worry it happenes to a lot of people. Your nose and mouth are both connected to your throat and that sudden rush of fluids from your throat can confuse your body, Its like when you laugh while eating and then you accidently get food in your nose. Don't worry this isn't anything that is particulary bad but i will say that the vomiting is not good

3 0

4 years ago

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An archer shoots an arrow at an 85.0 m distant target; the bulls eye of the target is at same height as the release height of th

inessss [21]

Answer:

a) 14.1°

b) over

Explanation:

The usual model of ballistic motion assumes that the only force on the flying object is that due to gravity. When an object is launched with initial velocity v0 at some angle θ with respect to the horizontal, the distance it travels is ...

d = (v0)²sin(2θ)/g

Using this relation, we can find the launch angle to make the object travel a given distance:

θ = 1/2arcsin(dg/v0²) . . . . where g is the acceleration due to gravity

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For the arrow to hit a target 85 m away at the same height it was launched with speed 42.0 m/s, the launch angle must be ...

θ = 1/2arcsin(dg/v0²) = 1/2(arcsin(85·9.8/42²)) ≈ 14.0893°

The arrow must be released at an angle of about 14.1°.

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The flight time to the tree at a distance of 42.5 m will be that distance divided by the horizontal speed:

t = 42.5/(42cos(14.0893°)) ≈ 1.0433 . . . . seconds

The height at that time is ...

h(t) = -4.9t² +42sin(14.0893°)t ≈ 5.33 . . . meters

The arrow will go <em>over</em> the branch.

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<em>Additional comment</em>

Since gravity provides the only force on the arrow, its horizontal speed is constant at vh = v0·cos(θ), when the arrow is launched with speed v0 at angle θ above the horizontal. Its vertical speed will be reduced by the acceleration of gravity, so will be vv = v0·sin(θ) -gt. The height is the integral of the vertical speed, so is ...

h(t) = (1/2)gt² +v0·sin(θ)t

The height will be 0 at t=0 and at t=2v0sin(θ)/g, so the horizontal distance traveled will be ...

d = vh·t

= (v0·cos(θ))(2v0·sin(θ)/g) = (v0²/g)(2·sin(θ)cos(θ))

= v0²sin(2θ)/g

Note that this is all simplified by the fact that the target and launch point are at the same level (h=0).

6 0

3 years ago

Question 1 Unsaved

aleksandrvk [35]

Sun is the biggest mass in the ss

4 0

3 years ago

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An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

1. (15 points) Small cart is rolling down an inclined track and accelerating. It fires a ball straight out of the cannon as it m

Elena L [17]

Answer:

Explanation:

This problem can be solved with the conservation of the momentum.

If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.

Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.

If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.

HOPE THIS HELPS!!

5 0

3 years ago