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tiny-mole [99]
3 years ago
10

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

traight flight path, which sets the wings in circular motion. If it takes it 35 s to complete the circle and the wingspan of the plane is 11 m, what is the acceleration of the wing tip?
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s

Acceleration

a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2

Acceleration of the tip of the plane is 0.17724 m/s²

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A 2.7 kg block of ice at a temperature of 0.0◦C and an initial speed of 6.5 m/s slides across a level floor. If 3.3 × 105 J are
bixtya [17]

Answer:

The mass of ice melted is: 1.728*10^{-4}(Kg)

Explanation:

We need to remember that the latent heat of fusion is related as:L_{f} =\frac{Q}{m}, where Lf is the latent heat of fusion, Q is the amount of heat and m is the mass, and when there is a change of phase, e.g ice into liquid water, the temperature remains constant during this process. Now we calculate the kinetic energy of the block as: E_{k}=\frac{m*v^{2} }{2}=\frac{2.7*6.5^{2} }{2} =57.02(Joules), so this energy is used to change ice into liquid water, and knowing L_{f}=3.3*10^{5}(Joules/Kg), then we can replace in the equation the latent heat of fusion and get the mass of ice melted as:m=\frac{L_{f} }{Q} =\frac{57.04}{3.3*10^{5} } =1.728*10^{-4}(Kg)

4 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
tamaranim1 [39]

Answer:

A: 5.67*10^-5 N

B: 3.49*10^-5 N

C: -9.16*10^-5 N

Explanation:

\\A\\F_{A} = F_{BA} + F_{CA} =G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(\frac{363*517}{0.5^2} + \frac{154*363}{0.75^2})\\\\= 5.67*10^-5 N \\\\B\\\\F_{B} = -F_{BA} + F_{CB} =-G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{b}}{r^2_{BC}} \\\\= (6.67*10^(-11))*(-\frac{517*363}{0.5^2} + \frac{517*154}{0.25^2})\\\\= 3.49*10^-5\\\\

F_{C} = -F_{CA} - F_{CB} =-G*\frac{m_{b}*m_{c}}{r^2_{BC}} - G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(-\frac{517*154}{0.25^2} - \frac{154*363}{0.75^2})\\\\= -9.16*10^-5

7 0
3 years ago
X
Gekata [30.6K]

Answer: is option C: <em>Prolonged periods of cooling and warming</em>.

Explanation:

In the history of Earth, climate varies time to time. At times, the Earth's atmosphere was much hotter and humid as compare to the present time, but similarly it has been noticed that climate also has been much colder than he present time, whereas the number of glaciers covers much of the Earth's surface. There are two kinds of periods in which we further classified Earth's climate namely, Glacial period, and Inter-glacial periods. It has been noticed that the average global temperature of Earth during glacial periods was around 5.5°C or 10°F, which is less than Earth's present climate. On the other hand, during inter-glacial periods Earth's temperature was about  1.1°C or 2.0°F, which is again higher as compared to current temperature. Over the past 900,000 years, Earth's temperature varied less than 5°C. Scientist believe by looking at the Earth's climate history, that glaciers will proceed again in formation, but it will take thousands of years.

8 0
3 years ago
in as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric
Pie

Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

I multiple 50 with 10 then  with 30 so I know how much watt the fan takes at June.

Since there are 2 light bulb I multiple 10 with 2 than with 8 than with 30.

15000 watts for the fan,

4800 watts for light bulb,

add them and then times it by 10.

Rs198000

4 0
3 years ago
Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving
Papessa [141]

Answer:

The final velocity of the object after 2 seconds is 30 m/s

Explanation:

Given;

constant downward acceleration, a =  10 m/s²

initial velocity of the object falling down, v = 10 m/s

time of fall, t = 2 s

The final velocity of the object is given by;

v = u + at

where;

v is the final velocity

v = 10 + (10)(2)

v = 10 + 20

v = 30 m/s

Therefore, the final velocity of the object after 2 seconds is 30 m/s

3 0
3 years ago
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