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ololo11 [35]
4 years ago
10

A sack of potatoes weighing 16.0-kg falls from a very tall building. At a certain point at the motion downwards, its measured ac

celeration is 4.6 m/s2 and its velocity is 54.2 m/s. Assuming that the magnitude of the drag force due to air resistance is proportional to the square of its speed, What is its terminal speed?
Physics
1 answer:
Vitek1552 [10]4 years ago
7 0

Answer:

The terminal speed is 74.833 m/s

Explanation:

The drag force is equal to square of speed:

Fdrag = k*v²

According Newton`s law:

Fnet = m*a

m*g - k*v² = m*a

k=\frac{m(g-a)}{v^{2} }

k=\frac{16*(9.8-4.6)}{54.2^{2} } =0.028

If terminal speed, the net force is zero.

kv_{t} ^{2} =mg\\v_{t} =\sqrt{\frac{mg}{k} } =\sqrt{\frac{16*9.8}{0.028} } =74.833m/s

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kiruha [24]
The object is not moving.

My explanation is that say if you sit a ball on the table and it is a smooth surface with no bumps or anything. The ball will sit still since it can’t roll unless you hit it.

Hope this helps!
5 0
3 years ago
Use Bernoulli's Equation to find out how fast water leaves an opening in a water tank. The water level is 0.75 m above the openi
yuradex [85]

Answer:

3.84 m/s

Explanation:

Using Bernoulli's equation below:

P1 + (1/2ρv1²) + h1ρg = P2 + (1/2ρv2²) + h2ρg

where P1 = P2 atmospheric pressure

(1/2ρv1²) + h1ρg = (1/2ρv2²) + h2pg

collect the like terms

h1ρg - h2ρg = (1/2ρv2²) - (1/2ρv1²)

factorize the expression by removing the like terms on both sides

gρ(h1 - h2) = 1/2ρ( v2² - v1²)

divide both side by rho (density in kg/m³, ρ )

g(h1 - h2) = 1/2 (v2² - v1²)

assuming the surface of the tank is large and the speed of water then at the tank surface  v1 = 0

2g(h1 - h2) = v2²

take the square root of both side and h1 - h2 is the difference between the surface of the tank and the opening where water is coming out in meters

√2g(h1 - h2) = √ v2²

v2 = √2g(h1-h2) = √ 2 × 9.81×0.75 = 3.84 m/s

6 0
3 years ago
a bag of cement is pulled along a horizontal plane by a constant force of 15n calculate the work done in moving the bag through
morpeh [17]

Answer:

300J

Explanation:

Given parameters:

Constant force = 15N

Distance  = 20m

Unknown:

Work done   = ?

Solution:

Work done is the force applied to move a body through a particular distance.

 Work done  = Force x distance

 So;

  Work done  = 15 x 20  = 300J

4 0
3 years ago
The position of a particle is given by the expression x = 2.00 cos (6.00πt + 2π/5), where x is in meters and t is in seconds det
Nana76 [90]
The position of the particle at time t is
x(t) = 2.00 \cos (6.00 \pi t+ \frac{2}{5}\pi )
If we substitute t=0.330 s, we find
x(0.330 s)=2.00 \cos (6.00 \pi (0.330 s) +  \frac{2}{5}  \pi)=
=2.00 \cos (1.98 \pi +  0.40 \pi )=2.00 \cos (2.38 \pi) =
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8 0
3 years ago
A proton in a linear accelerator has a de broglie wavelength of 156 pm . part a what is the speed of the proton?
SpyIntel [72]

To calculate the speed of the Proton we use, de Broglie equation as

\lambda =\frac{h}{mv}

Here, m is mass of proton and its value of 1.6726219 \times 10^{-27} kg and h is Plank constant and its value of 6.626\times 10^{-34} Js.

Given,  \lambda = 156 pm = 156 \times 10^{-12} m .

Substituting these values in above equation, we get

 156 \times 10^{-12} m = \frac{6.626\times 10^{-34} Js}{1.6726219 \times 10^{-27} kg\times v} \\\\ v=\frac{3.96\times 10^{-7} }{156 \times 10^{-12} } = 0.02538 \times 10^{5} \ m/s\\\\ v = 2538 \ m/s

Thus, the speed of proton is 2538 m/s.


5 0
3 years ago
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