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ELEN [110]
2 years ago
13

Write the dimensional formula of gravitational constant and specific heat capacity.​

Physics
1 answer:
Goryan [66]2 years ago
6 0

Answer:

SEE EXPLANATION

Explanation:

The dimensional formula of gravitational constant is given by,

M^-1 L^3 T^-2  

Where,  

M = Mass

L = Length

T = Time

Dimensional formula of Specific Heat Capacity =[ M^0L^2T^-2K^-1]

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Um caminhão de entregas transporta produtos de tamanho e peso elevados, o que requer o uso de máquina simples para facilitar a d
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2 years ago
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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2 years ago
A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w
melomori [17]

The  initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

h = vi - 0.5(9.8)(3²)

h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

h = 0 + 0.5(9.8)(1.5)²

h = 11.025 ----------- (2)

solve (1) and (2) together, to determine the initial velocity of the ball

11.025 = vi - 44.1

vi = 11.025 + 44.1

vi = 55.125 m/s

Thus, the  initial velocity of the ball is 55.125 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

8 0
1 year ago
All stars in a stellar cluster have roughly the same:___.
3241004551 [841]

All stars in a stellar cluster have roughly the same distance.

<h3>What coloration are celebrity clusters?</h3>

Open clusters have a tendency to be blue in color. They frequently include glowing gas and dust. The stars in an open cluster are young stars that all formed from the equal nebula. These warm blue stars are in an open cluster known as the Jewel Bo

<h3>Are stars in the identical cluster?</h3>

Though stellar associations are grouped in with megastar clusters, they're pretty a bit different. "Stellar associations are companies of tens to hundreds of stars that have similar a while and metallicities, and are moving in roughly the equal direction within the galaxy, but are no longer gravitationally bound," Geller said.

Learn more about star cluster here:

<h3>brainly.com/question/20326847</h3><h3 /><h3>#SPJ4</h3>
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1 year ago
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