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ycow [4]
4 years ago
12

An astronaut's pack weighs 17.5 N when she is on earth, but only 3.24 N when she is on the surface of an asteroid. What is the a

cceleration due to gravity on this asteroid?
9.21 m/s^2
1.43 m/s^2
1.81 m/s^2
1.28 m/s^2
5.40 m/s^2
Physics
1 answer:
Nimfa-mama [501]4 years ago
7 0

Answer:

a= 1.81 m/s²

Explanation:

Ratio of the weights on both surfaces can be calculated as,

\frac{W_{a} }{W_{E} }=\frac{ma}{mg}

Here;

   Wa is the weight on the asteroid,       W_{E} is weight on earth,

         m is mass of pack,          a is acceleration due to an asteroid and

         g is acceleration due to gravity.

         Rearrange above equation for a,

           a=g (\frac{Wa}{W_{E} })

Substitute 3.24 N for Wa,       17.5 N for  W_{E},

          9.81 m/s² for g in the above equation,

          a= 9.81(\frac{3.24}{17.5})

          a= 1.81 m/s²

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Can you plz answer it!!
LUCKY_DIMON [66]

Answer:

B

Explanation:

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Which of the following statements best describes energy conservation in heat engines?
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B is the correct answer
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3 years ago
A source emits sound with a frequency of 860 Hz. It is moving at 20.0 m/s toward a stationary reflecting wall. If the speed of s
Galina-37 [17]

Answer:

f_o=860\ Hz

Explanation:

Given:

  • original frequency of sound wave, f=860\ Hz
  • speed of the sound source, v_s=20\ m.s^{-1}
  • original speed of sound wave from the source, s=343\ m.s^{-1}

<u>According to the Doppler's effect:</u>

\frac{f_s}{f_o} =\frac{s+v_s}{s-v_o}

The sound is reflected from the wall and the source is moving towards the wall and observer is also riding the same source.

The velocity of the observer, v_o=-20\ m.s^{-1}

\frac{860}{f_o} =\frac{s+20}{s-(-20)}

f_o=860\ Hz

3 0
3 years ago
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Force X has a magnitude of 1260 ​pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4
weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530  P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= -  1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the  equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny=  -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2}  }

F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2}  }

Fe= 2579.68 P

Calculation of the direction of  equilibrant force (α)

\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )

\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )

α= 24.8°

Look at the attached graphic

6 0
3 years ago
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