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PolarNik [594]
4 years ago
10

An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.50 km. If the airplane rounds half

the circle in 1.50 3 102 s, determine the magnitude of its (a) displacement and (b) average velocity during that time. (c) What is the airplane’s average speed during the same time interval?
Physics
1 answer:
notsponge [240]4 years ago
4 0

Explanation:

A.

Displacement,S; Θ = S/r

= 180/360 * 2π * 3500

= 10995.6 m

Θ = 10995.6/3500

= 3.142

B.

Angular speed,w = Θ/t

= 3.142/1.503102

= 2.09 rad/s

Velocity = w * r

= 2.09 × 3500

= 7315 m/s.

C.

The same as B.

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If you say displacement is greater than distance, you will contradict the above statement. Displacement is always less than or equal to distance. Note that distance is a scalar whereas displacement is a vector.So' displacement cannot be more than distance.
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Each of three identical 15.0-L gas cylinders contains 7.50 mol of gas at 295 K. Cylinder A contains He, cylinder B contains O2,
Virty [35]

Answer:

Helium will have the highest average speed

Explanation:

the kinetic molecular theory tells us that the temperature of a gas is directly proportional to the average kinetic energy of the gas. So since all three gases have the same temperature, they must have the same kinetic energy. Kinetic energy is KE = 0.5mv². How then can a gas with such a small molar mass like helium have the same kinetic energy as much heavier gases like oxygen and sulphur dioxide? By having a much larger average speed

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3 years ago
A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And
Naily [24]

Answer:

700J

Explanation:

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3 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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Lena [83]

Answer:

6:50 PM

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