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Aneli [31]
3 years ago
11

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. assume that the plane reaches the required takeoff velocity at the end of the runway. what is the time tto needed to take off?
Physics
1 answer:
ElenaW [278]3 years ago
6 0
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
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strojnjashka [21]

speed of the plane is given as

v = 442 km/h = 122.8 m/s

height of the plane is

h = 575 m

acceleration due to gravity is

g = 9.80 m/s^2

part a)

when package is dropped its vertical speed is zero so we can use kinematics to find the time of drop

\delta y = v_y * t + \frac{1}{2}gt^2

575 = 0 + \frac{1}{2}*9.80*t^2

t = 10.83 s

Part b)

Horizontal distance moved by the package is given by

\delta x = v_x * t

\delta x = 122.8 * 10.83

\delta x = 1330.3 m

Part c)

final speed in x direction

v_x = 122.8 m/s

final speed in y direction

v_y = 9.8*10.83 = 106.13 m/s

so net speed as it hit the ground will be

v = \sqrt{v_y^2 + v_x^2}

v = \sqrt{122.8^2 + 106.13^2}

v = 162.3 m/s


5 0
3 years ago
Urgent please help me
Verizon [17]

1433 km

Explanation:

Let g' = the gravitational field strength at an altitude h

g' = G\dfrac{M_E}{(R_E + h)^2}

We also know that g at the earth's surface is

g = G\dfrac{M_E}{R_E^2}

Since g' = (2/3)g, we can write

G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)

Simplifying the above expression by cancelling out common factors, we get

(R_E + h)^2 = \dfrac{3}{2} R_E^2

Taking the square root of both sides, this becomes

R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E

Solving for h, we get

h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})

\:\:\:\:\:= 1433\:\text{km}

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3 years ago
A spring has a spring constant of 100 N/m, and the mass hanging from is is 0.71 kg. What is the period of the spring's motion?
Anuta_ua [19.1K]

k = spring constant of the spring = 100 N/m

m = mass hanging from the spring = 0.71 kg

T = Time period of the spring's motion = ?

Time period of the oscillations of the mass hanging is given as

T = (2π) √(m/k)

inserting the values in the above equation

T = (2 x 3.14) √(0.71 kg/100 N/m)

T = (6.28) √(0.0071 sec²)

T = (6.28) (0.084) sec

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Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold
arsen [322]

Answer:

Resistance of gold wire, R=1977 \times 10^3 ohm

Explanation:

In this question we have given

Density of gold, d=19.3\times 10^3 \frac{kg}{m^3}

resistivity of gold, r=2.44\times 10^{-8} ohm.m

Length of wire, L= 2.05 km

Temperature, T= 20^oC

We know that relation between volume and density is given as

Density= \frac{mass}{Volume}

Therefore, volume occupied by one gram gold is given as,

V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3.........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

V=\pi \times r^2 \times L......(2)

Here,

A= \pi \times r^2...........(3)

here A is the cross sectional area of cylendrical gold wire  

From equation 2 and 3

we got

V=A \times L...............(4)

on comparing equation 1 and equation 4, we got,

A \times L=5.181\times 10^{-8} m^3

A=\frac{5.181\times 10^{-8} m^3}{2050 m}

A=2.53\times 10^{-11}m^2

we know that resistance and resistivity are related by following formula,

Resistance = resistivity\times \frac{L}{A}................(5)

Put values of resistivity, A and L in equation 5, we got

R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}

R=1977 \times 10^3 ohm

Therefore resistance of gold wire, R=1977 \times 10^3 ohm

7 0
3 years ago
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