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sdas [7]
2 years ago
10

an airplane has a maximum velocity of 160km/h in still air. calculate its maximum velocity when it travels in air with a crosswi

nd of 30km/h
Physics
1 answer:
NNADVOKAT [17]2 years ago
4 0

Answer:

Velocity can be directly added or subtracted.

For example, if a boat has a velocity V in still water.

And now you put the boat in a river with a current that has a velocity V'

The total velocity of the boat in that river is just the addition of these two velocities.

Velocity in the river  = V + V'

Where the only tricky part is that the velocity is a vector, so you need to take in account the directions of each vector.

In this case, we have a plane with a maximum velocity of 160km, let's assume a direction for this velocity, let's say that is in the positive x-direction.

Then we can write the velocity in the vector form:

velocity  = (vel in x-axis, vel in y-axis)

The velocity of the plane can be written as:

v = (160km/h, 0)

Now we add a crosswind of 30km/h

crosswind means that it is perpendicular, then it acts on the y-axis.

Then the total velocity of the plane will be:

velocity = (160km/h, 0) + (0, 30km/h)

velocity = (160km/h, 30km/h)

Now you can compute the total velocity of the airplane as the module of that vector.

Remember that for a vector (x, y) the module is:

mod = √(x^2 + y^2)

Then the module of the velocity is:

v = √( (160km/h)^2 + (30km/h)^2) = 162.8 km/h

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If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu
zubka84 [21]

Answer:

minimum speed is 15.35 m/s

frictional coefficient  is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and   minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² =  236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 =  0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245  = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient  is 0.26

4 0
2 years ago
Read 2 more answers
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
atroni [7]

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

brainly.com/question/11049671

#SPJ4

4 0
1 year ago
Help! I don’t really know what it’s asking
Misha Larkins [42]
You have to do the math of each and see which one adds up to 66.5
6 0
2 years ago
Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
2 years ago
Is a sugar crystal a mineral? Explain
DENIUS [597]
No because sugar is made up of organic material 
5 0
2 years ago
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