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Daniel [21]
3 years ago
13

A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is

+72.2 m/s. (+ indicates to the right). Determine the average acceleration for the plane.
Physics
1 answer:
IrinaK [193]3 years ago
4 0

Answer:

Acceleration, a=2.48\ m/s^2

Explanation:

Given that,

The plane is at rest initially, u = 0

Final speed of the plane, v = 72.2 m/s

Time, t = 29 s

We need to find the average acceleration for the plane. It can be calculated as :

a=\dfrac{v-u}{t}

a=\dfrac{72.2}{29}

a=2.48\ m/s^2

So, the average acceleration for the plane is 2.48\ m/s^2. Hence, this is the required solution.

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True the plastic will float because of the principle of flotation or buoyancy

Explanation:

Buoyancy explains it all!!

Buoyancy  is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.

According to the principle of flotation:

<em>"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"</em>

The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

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3 years ago
At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t
STatiana [176]
I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
4 0
3 years ago
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5kg of copper ball and I kg of water are at the same temperature of 31 C. Which
blagie [28]

Answer:

r

Explanation:

8 0
3 years ago
Which of the following statements describes the movement of air?
Nutka1998 [239]
The movement of air flows from high pressure to low pressure
3 0
3 years ago
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different
kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

                           P = P_a* ( 1 - βh/T_a)^(g / R*β)

                          (g / R*β) = 32.174 / 1716*0.0035 = 5.252

                            P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

                            P = 1240 lb/ft^2

- The air density method which is expressed as:

                            P = P_a - γ*h

                            P = 2116.2 - 0.07647*14,110

                            P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

                            P = P_a* e^ ( -g*h / R*T_a )

                            P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

                            P = 1270 lb/ft^2

6 0
3 years ago
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