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natita [175]
4 years ago
5

In fighting forest fires, airplanes work in support of ground crewsby dropping water on the fires. A pilot is practicing by drop

ping acanister of red dye, hoping to hit a target on the ground below. Ifthe plane is flying in a horizontal path 90.0 m above the ground and with a speed of69.0 m/s (153 mi/h), at what horizontal distance from thetarget should the pilot release the canister?
Physics
1 answer:
alexira [117]4 years ago
7 0

Answer:295.71 m

Explanation:

Given

Plane is flying 90 m above

Speed of plane=69 m/s

Let x be the distance from the target pilot release the canister

time taken to travel 90 m is

h=ut+\frac{at^2}{2}

90=0+\frac{gt^2}{2}

t^2=18.367

t=4.28 s

x=ut

x=69\times 4.28=295.71 m

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A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Vinvika [58]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

4 0
4 years ago
A car has a kinetic energy of 432,000 J when traveling at a speed of 23 m/s. What is its mass?
gladu [14]

Answer:

2

Explanation:

2

8 0
3 years ago
Which system works similar to a refrigeration system? ANSWER: A
elixir [45]
Heat Pump is the answer
5 0
3 years ago
Read 2 more answers
If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a
Sidana [21]
The force acting in the front direction is the 130N.
The frictional force is acting backwards          30N.

1) The net force is 130N - 30N    =  100N

2)  s  = ut + (1/2)at^2             u = 0,  Start from rest,  s = 25m t =5.

25 = 0*5  +  (1/2)* a * 5^2.

25 = 0  +  25/2  * a.

25  =    (25/2)a.      Divide 25 from both sides.

1 =  (1/2)* a.          Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force,  F = ma
                   100 =  m*2.        Divide both sides by 2.
                    
                     100/2  =  m
                       50       =  m.
                        m  = 50 kg.

4)  Final velocity ,   v = u + at.
                                   v  =  0  + 2*5 = 10 m/s.
                                  
   Kinetic Energy,  K  =  (1/2) * mv^2.
                                    =    1/2  * 50 * 10 * 10.
                                    =      2500 J.
3 0
3 years ago
1.A car starts from rest and acquires a velocity of 54 km/h in 2 minutes.Find(i) the acceleration and(ii) distance travelled by
Sergeu [11.5K]

Answer:

1.) 1620 km/h^2

2.) 2.7 km

Explanation:

1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,

U = 0.

Final velocity V = 54 km/h

Time t = 2 minute = 2/60 = 1/30 hour

Acceleration a will be change in velocity per time taken. That is,

a = ( V - U )/ t

Substitute V, U and t into the formula

a = 54 ÷ 1/30

a = 54 × 30 = 1620 km/h ^2

2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.

S = Ut + 1/2at^2

Substitute all the parameters into the formula

S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2

S = 54/30 + 810 × 1/900

S = 54/30 + 810/900

S = (1620+810)/900

S = 2430/900

S = 2.7 km.

Therefore, distance travelled by the car during this time is 2.7 km

4 0
3 years ago
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