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vodomira [7]
3 years ago
6

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

the water remains in the container at the top of the circle. ANS. 3.1 m/s E-4, P-12?​
Physics
1 answer:
12345 [234]3 years ago
6 0

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

  v = √Rg

  v = √(1.0(9.8))

  v = 3.1304951...

  v = 3.1 m/s

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Rashid [163]

Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 m/s^2

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion s=ut +\frac{1}{2}gt^2, we have

4.14 = 7.4t + 0.5\times9.8t^2

4.9t^2 + 7.4t - 4.14 =0

Using the quadratic formula \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

t = 0.43 s

3 0
4 years ago
A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that fina
lisov135 [29]

Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

5 0
3 years ago
The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th
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According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
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4 years ago
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Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
Zanzabum

Calm, sunny days with wind moving away from the center.

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3 years ago
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Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th
vitfil [10]

Answer:

q=7.65*10^{-8}C

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C

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3 years ago
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