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Artyom0805 [142]
4 years ago
5

a. A lifted parcel of air will be colder (heavier) that the air surrounding it. Because of this fact, the lifted parcel will ten

d to sink back to its original position. b. The rate of adiabatic cooling or warming that occurs in a parcel of air that is unsaturated. c. The level at which a lifted parcel of air becomes saturated, marking the base of a cumuliform cloud d. ​ If a parcel of air expands and cools, or compresses and warms, and there is no interchange of heat with its outside surroundings. e. If rising air cools to its dew-point temperature, condensation results forming a cloud. Because heat added during condensation offsets some of the cooling due to expansion, the air cools at a lower rate. f. The level at which a lifted parcel of air in a conditionally unstable atmosphere becomes warmer (lighter) than its environment and continues to rise freely g. A lifted parcel of air will be warmer (lighter) than the air surrounding it, and thus will continue to rise upward, away from its original position.
Physics
1 answer:
kvv77 [185]4 years ago
5 0
I think the answer is A because it’s a better explanation
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AveGali [126]
Science
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4 0
3 years ago
Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and coll
gregori [183]

Answer:

a)  t = 3.027 10⁻⁹ s ,  b)   y = 2.25 10⁻² m

Explanation:

We can solve this problem using the kinematic relations

a) as on the x-axis there is no relationship

          vₓ = x / t

          t = x / vₓ

We reduce the magnitudes to the SI system

          x = 5.6 cm (1m / 100 vm) = 0.056 m

we calculate

          t = 0.056 / 1.85 10⁷

          t = 3.027 10⁻⁹ s

b) the time is the same for the two movements, on the y axis

         y = v₀t + ½ a t²

         

as the beam leaves horizontal there is no initial vertical velocity

         y = ½ a t²

         

let's calculate

         y = ½  5.45 10¹⁵ (3.027 10⁻⁹)²

         y = 2.25 10⁻² m

6 0
3 years ago
A baseball is thrown directly upward from ground level with a velocity of +15 m/s. What are the two times when the ball is 10 m
erastovalidia [21]

Answer:

time is 0.5660 s

and time is - 3.62431  s

Explanation:

velocity u = 15 m/s

height s = 10 m

acceleration due to gravity g =  –9.8 m/s²

to find out

time

solution

we will apply here distance equation that is

s = ut - 1/2× gt²   ...........1

here put all these value and get time t

here s is height and g is -9.8

so

s = ut - 1/2× gt²

10 = 15t - 1/2× (-9.8)t²

10 = 15t + 4.9t²

solve it we get t

t = 0.56630 and -3.62431

so time is 0.5660 s

and time is - 3.62431  s

8 0
4 years ago
Question 31 Unsaved
Bumek [7]
A) elements in the same group usually have similar chemical properties
B) This is because their atoms have the same number of electrons in the highest occupied energy shells
C) Lithium, sodium, potassium in group 1 elements are reactive metals called the alkali metals
7 0
3 years ago
A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

3 0
3 years ago
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