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ladessa [460]
3 years ago
12

on a recent adventure trip, Anita Break went rock climbing. Anita was able to perform 1370 J of work in 100 seconds. Determine A

nita's power rating during this portion of the climb
Physics
1 answer:
Zanzabum3 years ago
5 0

        Power = (energy) / (time)

         =    (1370 joules) / (100 seconds)

         =       13.7  joules/second

         =       13.7 watts .

That's not an awful lot of power, especially for a strenuous activity like
rock-climbing.  Shoot !  Even I could probably perform at that level.

Compare 13.7 watts to the light power coming out of a 20-watt night light.

         13.7 watts  =  0.018 horsepower.      (rounded)
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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
What is the approximate terminal velocity of a sky diver before the parachute opens
anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

  • The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
  • The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
  • During terminal velocity, we can represent mathematical equation as;

                           Buoyancy force + drag force = Gravity

6 0
3 years ago
This is for physical science. Can someone help me?
Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
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when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results
Nikitich [7]

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

  • Hard stabilization is the prevention of erosion through the use of artificial barriers.
  • Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
  • Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
  • These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
  • Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

brainly.com/question/16022736

#SPJ4

4 0
2 years ago
particle moves over three quarters of a circle of radius R what is the magnitude of its displacement​
fomenos

Answer:

is equal to threee idont know

Explanation:

hindi ko alam ang explanation kwa kwek kwen hahahahhahahahaha

sanay akoy pag padendyahan niyo nay. hahahahhahahaha maawa kayo saakin mga aking mahal hahahhahahahahhaha hahah en then go char hahahhaa

6 0
3 years ago
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