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vampirchik [111]
3 years ago
8

How many coulombs of positive charge are there in 0.1 kg of carbon? Twelve grams of carbon contain Avogadro's number of atoms, w

ith each atom having six protons and six electrons.
Physics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

481.84\times 10^{4}

Explanation:

We have given the mass of the carbon =0.1 kg

Molar mass of the carbon =12 gram

Number of moles n=\frac{0.1}{0.012}=8.333

We know that 1 mole of carbon contain 6.023\times 10^{23}atoms

So atoms contained by 8.333 moles =8.333\times 6.023\times 10^{23}=50.191\times 10^{23}atoms

In question it is given that each atom contain 6 proton so number of proton 6\times 50.191\times 10^{23}atoms=301.15\times 10^{23}proton

Now we know that each proton contain positive charge of 1.6\times 10^{-19}C

So total charge =1.6\times 10^{-19}\times 301.15\times 10^{23}=481.84\times 10^{4}C

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Efficiency =  Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

 = 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s =  0.061667 MW              J/s = Watts

Power input = 0.061667 MW = 61 667 W

From  Efficiency =  Power Output / Power Input

   28% =  Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W
 
  Power Output = 17 267 W

 Rate of heat Rejection = Power input - Power output

                                        = 61667 - 17267 = 44400 W

Rate of heat Rejection = 44 400 W.


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3 years ago
A mechanic uses a wrench that is 22 cm long to tighten a spark plug. If the mechanic exerts a force of 58 N to the end of the wr
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The maximum torque she can apply to the spark plug is 12.76 Nm.

<h3>What is torque?</h3>

Torque is the rotational equivalent of linear force obtained by taking the product of force and the radius of the applied force.

τ = Fr

where;

  • F is the applied force
  • r is the radius

The maximum torque she can apply to the spark plug is calculated as follows;

τ = 58 x (0.22)

τ = 12.76 Nm

Thus, the maximum torque she can apply to the spark plug is 12.76 Nm.

Learn more about torque here: brainly.com/question/14839816

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The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
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The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
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  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

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A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
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(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

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