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djverab [1.8K]
3 years ago
11

If a plane is traveling at a velocity of 100 km/hr East, and if the wind velocity is 25 km/hr to the East, then what is the velo

city of the plane relative to an
observer on the ground below?
100 km/hr East
75 km/hr West
75 km/hr East
125 km/hr East
Physics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

75 west

Explanation:

I think not sure lucky guess

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Help me with this question please
vichka [17]

Answer:

Its true i'm pretty sure

Explanation:

7 0
2 years ago
Read 2 more answers
Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
Sunspots are dark because ___________. a. the strong magnetic field inhibits the currents of hot gas rising from below. b. of sh
maxonik [38]

Answer:

a) the strong magnetic field inhibits the currents of hot gas rising from below.

Explanation:

Sunspots are dark because they are a little cooler than the photo sphere.

The magnetic field inhibits the strong rising currents of hot gas,allowing it to cool the surface.

4 0
3 years ago
Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0
3 years ago
Read 2 more answers
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