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ipn [44]
3 years ago
9

Write the equation of motion gover the centre of mass​

Physics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

write the equation of motion go over the centre of mass

Explanation:

the center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

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A force is a push or a pull. Under the right circumstances, anything can exert a force.
Sliva [168]

Answer:

Explanation:

the force that you applied caused the notebook to move.

the force applied on the notebook by the table causes it to stop moving.

this is because after sometime the book uses up the force and later the force you applied is less than that of the force by the table.

7 0
3 years ago
The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0
3 years ago
Read 2 more answers
Which of the following occurs with both a cold front and a mountain breeze?
BigorU [14]

Answer:

b warm air rises

Explanation:

5 0
3 years ago
Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.
RSB [31]

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

.

5 0
2 years ago
The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
schepotkina [342]

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
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