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eimsori [14]
3 years ago
15

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

he wind and the speed of the airplane in still air. g
Physics
1 answer:
Tamiku [17]3 years ago
5 0

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

V_a= Velocity of airplane in still air

V_w= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)

Subtracting the two equations we get

V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h

Applying the value of velocity of wind to the first equation

V_a-40=460\\\Rightarrow V_a =500\ km/h

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

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Ghella [55]

Answer:

Pressure on both feet will be 500N/m^2  

Explanation:

Weight of the person F = 500 N

Area of foot print A=0.5m^2

Area of both the foot A=2\times 0.5=1m^2

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure P=\frac{F}{A}

P=\frac{500}{1}=500N/m^2

So the pressure on both feet will be 500N/m^2 when person stands on both feet.

7 0
3 years ago
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
2 years ago
. Maria walked 1.5 miles south to her house in 0.5 hours. What is her speed in miles per hour?
Simora [160]

1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

v=\frac{d}{t}

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

v=\frac{1.5}{0.5}=3 mi/h

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

d=1.5 mi \cdot 1609 m/mi = 2414 m is the displacement in meters

t=0.5 h \cdot 3600 s/h =1800 s is the time taken in seconds

Substituting,

v=\frac{2414 m}{1800 s}=1.34 m/s

And the direction of the velocity is the same as the displacement, so it is south.

6 0
3 years ago
a man hikes 6.6 km north along a straight path with an average velocity of 4.2 km/h to the north. he rest at a bench for 15 min.
SSSSS [86.1K]

Answer:

2.6h

Explanation:

I attached the image below of the work hope you can see it. Hope this helps!

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What is used as evidence for sea-floor spreading?
raketka [301]

Answer:

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The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.

In previous theories,  continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves  as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.

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3 years ago
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