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2 years ago

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:

hope it helps

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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2 years ago

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Yuri [45]

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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.

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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

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3 years ago

Need Help With Physical Ed

Dimas [21]

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3 years ago

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What is atmospheric pressure​

What is atmospheric pressure