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nata0808 [166]
3 years ago
10

Anybody know how to get minimum wind velocity?

Physics
1 answer:
IRISSAK [1]3 years ago
8 0
Try looking at it like this and I'll bet you'll get it:

You're running through the boxcar on the train at 112 meters per second, but you notice that you're moving along the tracks at 210 meters per second. What must be the speed of the train along the tracks ?
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A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c
UkoKoshka [18]

Answer:0.00125 watts

Explanation:

resistance=50 ohms

Current=5 milliamps

Current=5/1000 milliamps

Current =0.005 amps

power=(current)^2 x (resistance)

Power=(0.005)^2 x 50

Power=0.005 x 0.005 x 50

Power=0.00125 watts

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3 years ago
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3 years ago
The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by
Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}

we have

a. for c = 20 cm

m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.

Evaluated at c = 20 cm, we have

\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0
3 years ago
What makes an electric charge?
scZoUnD [109]

Answer:

Most electric charge is carried by the electrons and protons within an atom. Conversely, two protons repel each other, as do two electrons. Advertisement. Protons and electrons create electric fields, which exert a force called the Coulomb force, which radiates outward in all directions.

5 0
3 years ago
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Two vehicles A and B accelerate uniformly from rest.
spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

5 0
3 years ago
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