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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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In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8
BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

      F = m a

      a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

       P = I / c        absorbent surface

       P = F / A

       F / A = I / c

       F = I A / c

The area of ​​area of ​​a circle is

      A = π r²

We replace

     F = I π r² / c

Let's calculate

     F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

     F = 8.375 10⁻²³ N

Density is

      ρ = m / V

      m = ρ V

      m = ρ (4/3 π r³)

      m = 4500 (4/3 π (1 10⁻⁶)³)

      m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

     a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

     a = 4.44 10⁻⁹ m/s²               absorbent surface

The correct answer is B

4 0
3 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
3 years ago
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USPshnik [31]
To find the scientific notation, you need to divide at the decimal by the power of 10. So since there are 2 powers of 10, what you want to do is move the decimal 2 places to the left which will give you: .054
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Kipish [7]

Answer:

235

Explanation:

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These techniques are central to psychoanalytic therapy. They can be used alone or in combination with one another. Their purpose is to increase awareness and foster insight into the client's behavior and emotions

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