What is a normal force?

Assume that the complete combustion of one mole of ethanol to carbon dioxide and water liberates 1370 kJ of energy (Δ????°′=−137

Nikitich [7]

Answer:

28 moles of X.

Explanation:

Let's consider the complete combustion of ethanol.

C₂H₆O(l) + 3 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)

When 1 mole of ethanol burns it releases 1370 kJ, that is, the enthalpy of combustion is -1370 kJ/mol. If the energy released by one mole of etanol were used to syntesize the compound X, it would take 45.9 kJ to form 1 mole of X, that is, the enthalpy of formation is -49.5 kJ/mol. Then,

$1mol(etanhol).\frac{-1370kJ}{mol(etanhol)} .\frac{1mol(X)}{-49.5kJ} =28mol(X)$

8 0

4 years ago

g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin

Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

$i.e Normal \ Force(N) = mg cos \theta$

And $\theta = 90$ so

$N = 0$

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work

Explanation:

5 0

3 years ago

Read 2 more answers

A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 45.0 deg

Romashka [77]

Answer:

a. 139.748J

b.3.486m/s

c.zero

Explanation:

a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°

Potential energy during release is calculated as: $PE=mgh$

#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.

Find the height when the child is pulled back:

$cos 45\textdegree=y/2.10\\\\y=1.48m$

#therefore,vertical difference is 2.1m-1.48m=0.62m

$\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J$

#Hence the potential energy during release is 139.748J

b. From a, above, we have PE=139.748J, M=23.0kg.

At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:

$KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s$

#Hence the velocity at the bottom of the swing is 3.486m/s

c. Work is calculated as the product of force by distance.

From a, b above we have mass as 23.0kg .

-since the distance of the ropes remained constant the change in distance is zero:

$W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0$

Therefore the work in the ropes is 0,zero.

8 0

3 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

A car traveling at a speed of 25 m/s increases its speed to 30 m/s in 10 sec. What is the acceleration of the car?

inysia [295]

Answer:

.5m/s/s

Explanation:

V + at

8 0

3 years ago