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Flauer [41]
2 years ago
5

Una bola de ping pong de 10 g. de masa rueda a 10 m/s hacia una bola de pool de 250 g de masa, inicialmente quieta. Luego del ch

oque la pelota rebota hacia la izquierda con una velocidad de 5 m/s. Calcular la velocidad que adquiere la bola de pool, si esta pelota se movía hacia la izquierda con una velocidad de 0.4 m/s antes del choque
Physics
1 answer:
Leni [432]2 years ago
3 0
Shahehehebeidvejdvdje xidveidbdhx rudhejdbdhd
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Please help ASAP! Willing to give brainliest.
poizon [28]

Answer: B

Explanation:the voltage is just like the force that drives the current through out the circui... When trippled, the force increases and the current increases since the resistance in the circuit remains constant.

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3 years ago
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A 1,120 kg car is travelling with a speed of 40 m/s. Find its energy.
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Answer:

this vehichle has 896,000 jules of energy

Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity

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11 Design Imagine that a scientistdiscovered a way to make africtionless surface. What wouldbe some useful applications forthis
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You could move something across the Earth with a little push. It would make fuel really efficient on those pathways. You could make a floor that is impossible to walk on. Everybody would just fall without traction.

Explanation:

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2 years ago
List and Explain three ways study groups benefit your learning
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The answer

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Thinking together, Better friendship, Makes teacher happy.

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You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
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