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2 years ago

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,

1 answer:

3 0

Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.

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For an object to be in static equilibrium, the net force must be equal to zero and and The center of gravity must be above the s

iren [92.7K]

Answer:

The normal force must equal the weight.

8 0

3 years ago

If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact w

Anarel [89]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

5 0

3 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.

3 0

3 years ago

What are some examples and non-examples of volume

Jlenok [28]

Answer:

Explanation:

Some correct non-examples are: A glass half-empty; Anything in two dimensions; The amount that covers something.

4 0

3 years ago

Read 2 more answers

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi

77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0

3 years ago