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Dahasolnce [82]
3 years ago
14

A 20.0-kg package is dropped from a high tower in still air and is "tracked" by a radar system. When the package is 25 m above t

he ground, the radar tracking indicates that its acceleration is 7.0 m/s2. Determine the force of air resistance on the package.
Physics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

56 N

Explanation:

There are only two forces acting on the package:

- The force of gravity, directed downward, equal to

mg

where m is the mass of the package and g is the acceleration due to gravity

- The air resistance, acting upward, let's label it with R

According to Newton's second law, the resultant of the two forces must be equal to the product between the mass, m, and the acceleration, a:

mg-R=ma

where we have:

m = 20.0 kg

g = 9.8 m/s^2

a = 7.0 m/s^2 is the acceleration when the package is at 25 m above the ground.

Substituting into the equation, we can find the magnitude of the air resistance at that altitude:

R=mg-ma=m(g-a)=(20.0 kg)(9.8 m/s^2-7.0 m/s^2)=56 N

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A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock do
Pachacha [2.7K]

Answer:

A) v = 28.3 m/s

B) t =  4.64 s

Explanation:

A)

  • Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

        v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h  (1)

  • Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

       \Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)

  • So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
  • Δh = 26.0 m + 14. 8 m = 40.8 m (3)
  • Replacing now in (1), we can solve for vf, as follows:

       v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)

B)

  • In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
  • 1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
  • 2) Time elapsed from this point until it hits the street, with vo=0.
  • For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

       v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)

  • Replacing by the givens in (5) and solving for Δt, we get:

       \Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)

  • For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

       \Delta h = \frac{1}{2} * g * t^{2}  (7)

  • Replacing by the givens and solving for t in (7), we get:

       t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)

  • So, total time is just the sum of (6) and (8):
  • t = 2.9 s + 1.74 s = 4.64 s
5 0
2 years ago
Assuming the average rate of heat energy flowing outwards from earth to be 0.063 W/m2 calculate the energy available at a hot sp
MaRussiya [10]

Answer:

"13.48 Kwhr" is the right solution.

Explanation:

The given values are:

Average rate of heat energy,

= 0.063 W/m²

Diameter,

= 8m

Efficiency of conversion,

= 50%

Now,

The area of hotspot will be:

⇒  A=\frac{\pi}{4} d^2

On substituting the values, we get

⇒      =\frac{3.14}{4} (8)^2

⇒      =0.785\times 64

⇒      =50.24 \ m^2

Total heat generation rate will be:

⇒  Q=q\times A

        =0.063\times 50.24

        =3.16 \ W

hence,

The electricity generation capacity will be:

⇒  P=\eta Q t

On substituting the values, we get

⇒      =0.5\times 3.16\times 8760

⇒      =13840.8 \ Whr

On converting into Kwhr, we get

⇒      =13.84 \ Kwhr

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Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your
forsale [732]

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

\\ \sf\longmapsto F_L=\mu sN

u s is coefficient of static friction and N is normal reaction

Or

\\ \sf\longmapsto F_L=\mu smg

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Answer: quarantine

Explanation:

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Answer:

9 lần

Explanation:

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