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vichka [17]
2 years ago
10

What is atmospheric pressure​

Physics
2 answers:
Kitty [74]2 years ago
8 0

Answer:

atmospheric pressure is the pressure exerted by the weight of the atmosphere on the earth surface. the pressure varies from place to place and also from time to time.

Explanation:

<em><u>hope </u></em><em><u>it </u></em><em><u>helps</u></em>

Mekhanik [1.2K]2 years ago
5 0

Answer:

the pressure within the atmosphere of Earth

Explanation:

Atmospheric pressure is a force in an area pushed against a surface by the weight of the atmosphere of the earth.

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lutik1710 [3]

To perform a drug lookup to ensure that the new compound has been added to the computer system properly we have to select the new from the toolbar at the top of the screen

<h3>What is a computer system?</h3>

A computer system is a collection of computers, related hardware, and related software. The central processing unit (CPU), memory, input/output, and storage devices are the four main components of a computer system. To produce the desired result, all of these parts operate in concert as a single unit.

Selecting the new from the toolbar at the top of the screen will allow us to run a drug lookup to make sure the new compound has been properly added to the computer system.

Therefore the correct answer is the option C

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3 0
2 years ago
39. I also use it when I do the homework. (par 1 line 3).
Yuri [45]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a
Angelina_Jolie [31]

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

l=\sqrt{(33)^2+(4.5)^2}

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

Vehicle\ displacement= d\cos\theta

Where, \theta is angle between rope and ground

d = pulled length of rope

Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}

Vehicle\ displacement=9.90\ feet

Hence, The vehicle displacement is 9.90 feet.

3 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
Need Help With Physical Ed
Dimas [21]

which of the following is not a barrier to physical activity it is fear of injury I think.

8 0
3 years ago
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