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astraxan [27]
4 years ago
8

Spinning satellites are particularly stable. suppose that the astronauts launching a particular satellite decide to increase its

angular velocity by a factor of 5. how will that change affect the satellite's angular momentum?
Physics
1 answer:
Crank4 years ago
3 0
Supposing velocity is speed and direction, and momentum is mass*velocity, if the velocity increases by a factor of 5, then so should the momentum regardless of the whole space thing.
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An airplane flies 120 km at a constant altitude in a direction 30.0° north of east. a wind is blowing that results in a net hori
astraxan [27]
 Best Answer:<span>  </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN 
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km 
The work done by the aircraft overcoming the wind is 
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N 
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
6 0
3 years ago
Read 2 more answers
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t
stiks02 [169]

Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will  be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

8 0
3 years ago
Read 2 more answers
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
3 years ago
A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface
valentinak56 [21]

Answer:

\frac{kQ}{r^2} r^

Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

where r is the unit vector in the direction of unit charge

E= \frac{kQ}{r^2} r^

4 0
3 years ago
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