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Bond [772]
2 years ago
12

When a jet plane is cruising at high altitude, the flight attendants have more of a "hill" to climb as they walk forward along t

he aisle than when the plane is cruising at a lower altitude. why does the pilot have to fly with a greater "angle of attack" at high altitude than at low?
Physics
1 answer:
anygoal [31]2 years ago
3 0

Answer:

Probably because the Bernoulli effect (lift) is insufficient in thinner air to keep the plane aloft - increasing the angle of attack will increase the lift on the airplane

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Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. W
bogdanovich [222]

Answer:

The position on the y-axis where the magnetic field is zero is at y = 10 cm

Explanation:

The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by

B = μ₀i/2πR

Now, let y be the point where the magnetic fields of both wires are equal.

So, the magnetic field due to wire 1 carrying current i₁ = 73 A is

B₁ = μ₀i₁/2π(y - 3) and

the magnetic field due to wire 2 carrying current i₂ = 31 A is

B₂ = μ₀i₂/2π(13 - y)

At the point where the magnetic field is zero, B₁ = B₂. So,

μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)

cancelling out μ₀ and 2π, we have

i₁/(x - y) = i₂/(13 - y)

cross-multiplying, we have

(13 - y)i₁ = (y - 3)i₂

Substituting the values of i₁ and i₂, we have

(13 - y)73 = (y - 3)31

949 - 73y = 31y - 93

Collecting like terms, we have

949 + 93 = 73y + 31y

1042 = 104y

dividing through by 104, we have

y = 1042/104

y = 10.02 cm

y ≅ 10 cm

So, the position on the y-axis where the magnetic field is zero is at y = 10 cm

6 0
3 years ago
you are driving along a country road when you suddenly notice a log in the road ahead of you and immediately apply your brakes.
DiKsa [7]

a. The car will not hit the tractor.

b. The car would travel a distance of 52.07 m before stopping

c. At the moment when the car stopped, the tractor is 11.5 m in front of the car.

Linear motion

From the question, we are to determine of you will hit the tractor before you stop

First, we will determine the time it will take the car to stop

From the given information,

Initial velocity, u = 27.0 m/s

a = -7 m/s² (Negative sign indicates deceleration)

v = 0 m/s (Since the car will come to stop)

From one of the equations of linear motion,

v = u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration

and t is the time taken

Putting the parameters into the equation, we get

0 = 27 + (-7)t

7t = 27

t = 27/7

t = 3.857 secs

This is the time it will take the car to stop

Now, we will determine the distance the car would travel after applying the brakes

Using the formula

S = (u + v) / t

Where S is the distance traveled

S = [(27 + 0) / 2] * 3.857

S = 13.5 * 3.857

S = 52.0695 m

S ≅ 52.07 m

This means the car would travel 52.07 m after applying the brakes

Now, we will determine the distance the tractor would have traveled when the car came to a stop

Speed of the tractor = 10.0 m/s

Time taken for the car to stop = 3.857 secs

Using the formula,

Distance = Speed × Time

Distance = 10.0 × 3.857

Distance = 38.57 m

Now, we will determine the distance between the car and the tractor when the car finally stopped

Distance between the car and the tractor = Distance ahead + Distance traveled by the tractor after the car stopped - Distance traveled by car after applying the brakes

Distance between the car and the tractor = 25 m + 38.57 m - 52.07 m

Distance between the car and the tractor = 11.5 m

Therefore, the distance the tractor was ahead of the car + the distance the tractor traveled after the car stopped is more than the distance the car traveled after applying the brakes (25 m + 38.57 m > 52.07), the car will not hit the tractor.

To know more about distance, refer: brainly.com/question/10903482

#SPJ4

<u><em>[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tractor 25.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. you immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2 .</em></u>

<u><em>a.will you hit the tractor before you stop?</em></u>

<u><em>b.how far will you travel before you stop or collide with the tractor?</em></u>

<u><em>c.if you stop, how far is the tractor in front of you when you finally stop?]</em></u>

7 0
1 year ago
Your car is skidding to a stop from a high speed. name all the forces that apply
DerKrebs [107]
Normal force, weight, Kinetic friction, and air resistance are a few I think of the top of my head.
I hope this helps
6 0
3 years ago
Upper A 12​-pound box sits at rest on a horizontal​ surface, and there is friction between the box and the surface. One side of
Maurinko [17]

Answer:

The coefficient of static friction is : 0.36397

Explanation:

When we have a box on a ramp of angle \alpha , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.

In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

N=m*g*cos(\alpha)

and the force of static friction (f) is given as the static coefficient of friction (\mu) times the normal N:

f=\mu *m*g*cos(\alpha)

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397

6 0
3 years ago
Waste reduction + rationalization of consumption + reuse + recycling / ways of preserving natural resources
forsale [732]

Answer:

Hello your question is vague hence I will provide a general answer on the importance of : Waste reduction, rationalization of consumption, reuse, recycling as ways of preserving the natural resources

answer :

Waste reduction : when we reduce the amount of wastage on items we use especially items produced from natural raw materials we will help preserve the natural resource because they can be used to produce varieties of other items

Reuse and recycling of waste products help keep our natural environment healthy thus preserving our natural resources.

Explanation:

Waste reduction, rationalization of consumption, reuse and recycling are all ways of preserving our natural resources

Waste reduction : when we reduce the amount of wastage on items we use especially items produced from natural raw materials we will help preserve the natural resource because they can be used to produce varieties of other items

Reuse and recycling of waste products help keep our natural environment healthy thus preserving our natural resources.

5 0
3 years ago
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