Ask question

Ask question

All categories

3 years ago

15

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

The speed of the plane relative to the air (called the air speed) as measured by instruments aboard the plane is 310 km/hour . In which direction should the pilot head?

1 answer:

Alex17521 [72]3 years ago

6 0

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

You might be interested in

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

Read the passage.

nevsk [136]

Answer:

d in my opinion is the most opitmal answer

Explanation:

aka a graph comparing distances traveled by objects thrown on Earth and the moon

7 0

3 years ago

Read 2 more answers

Which is MOST LIKELY a solid at room temperature?

Art [367]

I’m guessing b because hydrogen is in your room and maybe eliminate to o but it also can be D

my final is: B!

6 0

3 years ago

Read 2 more answers

Two identical balls are thrown vertically upward. The second ball is thrown with an initial speed that is twice that of the firs

9966 [12]

Answer:

The maximum height of ball 2 is 4 times that of ball 1

Explanation:

We can find the maximum height of each ball by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (we take upward as positive direction)

s is the displacement

At the maximum height, s = h and v = 0 (the final velocity is zero), so re-arranging the equation:

$h=\frac{-u^2}{2g}$

The first ball is thrown with initial velocity $u_1$ , so it reaches a maximum height of

$h_1 = -\frac{u_1^2}{2g}$ (the quantity will be positive, since g is negative)

The second ball is thrown with initial velocity

$u_2 = u_1$

so it will reach a maximum height of

$h_2 = - \frac{u_2^2}{2g}=-\frac{(2u_1)^2}{2g}=4(-\frac{u_1^2}{2g}) = 4h_1$

So, its maximum height will be 4 times the maximum height reached by ball 1.

8 0

3 years ago

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b

lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

v² = u² + 2as

Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m

3 0

3 years ago