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3 years ago

Air near Earth's surface rises up into the atmosphere as it is heated by _____.

Physics

2 answers:

cricket20 [7]3 years ago

8 0

<span>Air near Earth's surface rises up into the atmosphere as it is heated by "The Wind"

Hope this helps!</span>

chubhunter [2.5K]3 years ago

4 0

Answer:

a. the wind

Explanation:

Air near Earth's surface rises up into the atmosphere as it is heated by the wind. Then, convection transfers heat energy from warmer regions near the Earth's surface to regions higher up in the atmosphere, that happen because warmer air has a lower density than cold air

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A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

A jet aircraft with a mass of 4,475 kg has an engine that exerts a force (thrust) equal to 60,800 N. (a) What is the jet's accel

BartSMP [9]

Answer:

a) 13.59 m/s²

b) 67.95 m/s

c) 169.875 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{60800}{4475}\\\Rightarrow a=13.59\ m/s^2$

Acceleration of the jet is 13.59 m/s²

$v=u+at\\\Rightarrow v=0+13.59\times 5\\\Rightarrow v=67.95\ m/s$

Velocity attained at 5 seconds is 67.95 m/s

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{67.95^2-0^2}{2\times 13.59}\\\Rightarrow s=169.875\ m$

Distance traveled in the 5 seconds is 169.875 m

4 0

3 years ago

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Which statement demonstrates that ultraviolet (UV) rays are electromagnetic waves? A. They require a medium to travel. B. They d

Sauron [17]

B. They don't require a medium to travel.

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3 years ago

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A 4000 kg rocket is launched, shooting 50 kg of burned fuel from its exhaust at a velocity of -625 m/s. What is the velocity of

AnnZ [28]

Hopefully this will help you.

8 0

4 years ago

Help plss i just need it plss

Travka [436]

Answer:

ooh thanks but don't give me advise

4 0

3 years ago