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3 years ago

Why don't we feel pressure exerted by atmosphere?

1 answer:

ozzi3 years ago

6 0

Crushing pressure. Human bodies are used to air pressure. The air pressure in our lungs, ears and stomachs is the same as the air pressure outside of our bodies, which ensures that we don't get crushed. Our bodies are also flexible enough to cope when the internal and external pressures aren't exactly the same.

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Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe

nignag [31]

Answer:

a) x = 40 t , y = 39 t , z = 6 + 32 t - 16 t ², b) x = 80 feet , y = 78 feet , the ball came into the field

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

x₀ = 0

y₀ = 0

z₀ = 6 feet

x-axis (towards end zone, GOAL zone)

x = xo + v₀ₓ t

x = 40 t

y-axis (field width)

y = y₀ + $v_{oy}$ t

y = 39 t

z axis (vertical)

z = z₀ + v_{oz} t - ½ g t²

z = 6 + 32 t - ½ 32 t²

z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

z = 6

6 = 6 + 32 t - 16 t²

(t - 2)t = 0

t=0 s

t= 2 s

The ball position

x = 40 2

x = 80 feet

y = 39 2

y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

x_total = 150 feet

y _total = 80 feet

so we can see that the ball came into the field

6 0

3 years ago

The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21

andrezito [222]

Answer:

Explanation:

Given

average speed of train $(v_{avg})=216 kmph\approx 60 m/s$

Maximum acceleration=0.05g

Now centripetal acceleration is

$a_c=\frac{v^2}{r}$

$0.05\times 9.8=\frac{60^2}{r}$

r=7346.93 m

(b)Radius of curvature=900 m

therefore $a_c=\frac{v^2}{r}$

$v=\sqrt{a_cr}$

$v=\sqrt{0.05\times 9.8\times 900}$

$v=\sqrt{441}=21 m/s$

8 0

3 years ago

Can someone please help

coldgirl [10]

Answer:

Making a quick cut left to intercept a pass

Explanation:

It takes more energe to do than running

7 0

3 years ago

Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi

leonid [27]

Answer:

Dark matter does not affect our view, humans can see through them.

Explanation:

They do not affect our view because we can see right through the (weakly interacting) dark matter, as they do not interact or interfere with electromagnetic force.

Dark matter are often invisible substances and are difficult to spot because they don't absorb or reflect light.

7 0

3 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

t= 1hr

4 0

3 years ago

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