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Mbefore*Vbefore=Mafter*Vafter
Lesson 2e (3/8)
As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. InIn an explosion, momentum is conserved.
{http://www.physicsclassroom.com/Class/momentum/u4l2e1.gif} p2Momentum conservation is often demonstrated in a Physics class with a homemade cannon demonstration. A homemade cannon is placed upon a cart and loaded with a tennis ball. The cannon is equipped with a reaction chamber into which a small amount of fuel is inserted. The fuel is ignited, setting off an explosion that propels the tennis ball through the muzzle of the cannon. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball.
{http://www.physicsclassroom.com/Class/momentum/u4l2e3.gif} p3In the exploding cannon demonstration, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved.
p4As another demonstration of momentum conservation, consider two low-friction carts at rest on a track. The system consists of the two individual carts initially at rest. The total momentum of the system is zero before the explosion. One of the carts is equipped with a spring-loaded plunger that can be released by tapping on a small pin. The spring is compressed and the carts are placed next to each other. The pin is tapped, the plunger is released, and an explosion-like impulse sets both carts in motion along the track in opposite directions. One cart acquires a rightward momentum while the other cart acquires a leftward momentum. If 20 units of forward momentum are acquired by the rightward-moving cart, then 20 units of backwards momentum is acquired by the leftward-moving cart. The vector sum of the momentum of the individual carts is 0 units. Total system momentum is conserved.
{http://www.physicsclassroom.com/Class/momentum/u4l2e4.gif}
p5Equalconserve.
Equal and Opposite
Just like in collisions, the two objects involved encounter the same force for the same amount of time directed in opposite directions. This results in impulses that are equal in magnitude and opposite in direction. And since an impulse causes and is equal to a change in momentum, both carts encounter momentum changes that are equal in magnitude and opposite in direction. If the exploding system includes two objects or two parts, this principle can be stated in the form of an equation as:
{http://www.physicsclassroom.com/Class/momentum/u4l2e2.gif} If the masses of the two objects are equal, then their post-explosion velocity will be equal in magnitude (assuming the system is initially at rest). If the masses of the two objects are unequal, then they will be set in motion by the explosion with different speeds. Yet even if the masses of the two objects are different, the momentum change of the two objects (mass • velocity change) will be equal in magnitude.
p6The diagram below depicts a variety of situations involving explosion-like impulses acting between two carts on a low-friction track. The mass of the carts is different in each situation. In each situation, total system momentum is conserved as the momentum change of one cart is equal and opposite the momentum change of the other cart.
{http://www.physicsclassroom.com/Class/momentum/u4l2e5.gif}
In each of the above situations, the impulse on the carts is the same - a value of 20 kg•cm/s (or cN•s). Since the same spring is used, the same impulse is delivered. Thus, each cart encounters the same momentum change in every situation - a value of 20 kg•cm/s. For the same momentum change, an object with twice the mass will encounter one-half the velocity change. And an object with four times the mass will encounter one-fourth the velocity change.
p7SolvingSolving Explosion Momentum physics experiences. Consider for instance the following problem:
A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. The cannon is at rest when it is ignited. Immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in 0.0218 seconds. Determine the post-explosion speed of the cannon and of the tennis ball.
Like any problem in physics, this one is best approached by listing the known information.
Given:
||< Cannon: ||< m = 1.27 kg ||< d = 6.1 cm ||< t = 0.0218 s ||
Ball:
m = 56.2 g = 0.0562 kg
The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds. The speed can be assumed constant since the problem states that it was measured after the impulse of the explosion when the acceleration had ceased. Since the cannon was moving at constant speed during this time, the distance/time ratio will provide a post-explosion speed value.
vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded)
The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. If momentum is to be conserved, then the after-explosion momentum of the system must be zero (since the pre-explosion momentum was zero). For this to be true, then the post-explosion momentum of the tennis ball must be equal in magnitude (and opposite in direction) of that of the cannon. That is,
mball • vball = - mcannon • vcannon
The negative sign in the above equation serves the purpose of making the momenta of the two objects opposite in direction. Now values of mass and velocity can be substituted into the above equation to determine the post-explosion velocity of the tennis ball. (Note that a negative velocity has been inserted for the cannon's velocity.)
(0.0562 kg) • vball = - (1.27 kg) • (-280 cm/s)
vball = - (1.27 kg) • (-280 cm/s) / (0.0562 kg)
vball = 6323.26 cm/s
vball = 63.2 m/s
Using momentum explosion, the ball is propelled forward with a speed of 63.2 m/s - that's 141 miles/hour!
It's worth noting that another method of solving for the ball's velocity would be to use a momentum table similar to the one used previously in Lesson 2 for collision problems. In the table, the pre- and post-explosion momentum of the cannon and the tennis ball. This is illustrated below.
MomentumBefore Explosion
MomentumAfter Explosion
Cannon
0
(1.27 kg) • (-280 cm/s)= -355 kg•cm/s
Tennis Ball
0
(0.0562 kg) • v
Total
0
0
The variable v is used for the post-explosion velocity of the tennis ball. Using the table, one would state that the sum of the cannon and the tennis ball's momentum after the explosion must sum to the total system momentum of 0 as listed in the last row of the table. Thus,
-355 kg•cm/s + (0.0562 kg) • v = 0
Solving for v yields 6323 cm/s or 63.2 m/s - consistent with the previous solution method.
Using the table means that you can use the same problem solving strategy for both collisions and explosions. After all, it is the same momentum conservation principle that governs both situations. Whether
Whether it is

The momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in an isolated system, there are no exceptions to this law.
Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.
in velocity.
Mbefore*Vbefore=Mafter*Vafter
Lesson 2e (3/8)
As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the system (that is often a collection of fragments from the original object) have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved.
{http://www.physicsclassroom.com/Class/momentum/u4l2e1.gif} p2Momentum conservation is often demonstrated in a Physics class with a homemade cannon demonstration. A homemade cannon is placed upon a cart and loaded with a tennis ball. The cannon is equipped with a reaction chamber into which a small amount of fuel is inserted. The fuel is ignited, setting off an explosion that propels the tennis ball through the muzzle of the cannon. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball.
{http://www.physicsclassroom.com/Class/momentum/u4l2e3.gif} p3In the exploding cannon demonstration, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved.
p4As another demonstration of momentum conservation, consider two low-friction carts at rest on a track. The system consists of the two individual carts initially at rest. The total momentum of the system is zero before the explosion. One of the carts is equipped with a spring-loaded plunger that can be released by tapping on a small pin. The spring is compressed and the carts are placed next to each other. The pin is tapped, the plunger is released, and an explosion-like impulse sets both carts in motion along the track in opposite directions. One cart acquires a rightward momentum while the other cart acquires a leftward momentum. If 20 units of forward momentum are acquired by the rightward-moving cart, then 20 units of backwards momentum is acquired by the leftward-moving cart. The vector sum of the momentum of the individual carts is 0 units. Total system momentum is conserved.
{http://www.physicsclassroom.com/Class/momentum/u4l2e4.gif}
p5Equal and Opposite Momentum Changes
Just like in collisions, the two objects involved encounter the same force for the same amount of time directed in opposite directions. This results in impulses that are equal in magnitude and opposite in direction. And since an impulse causes and is equal to a change in momentum, both carts encounter momentum changes that are equal in magnitude and opposite in direction. If the exploding system includes two objects or two parts, this principle can be stated in the form of an equation as:
{http://www.physicsclassroom.com/Class/momentum/u4l2e2.gif} If the masses of the two objects are equal, then their post-explosion velocity will be equal in magnitude (assuming the system is initially at rest). If the masses of the two objects are unequal, then they will be set in motion by the explosion with different speeds. Yet even if the masses of the two objects are different, the momentum change of the two objects (mass • velocity change) will be equal in magnitude.
p6The diagram below depicts a variety of situations involving explosion-like impulses acting between two carts on a low-friction track. The mass of the carts is different in each situation. In each situation, total system momentum is conserved as the momentum change of one cart is equal and opposite the momentum change of the other cart.
{http://www.physicsclassroom.com/Class/momentum/u4l2e5.gif}
In each of the above situations, the impulse on the carts is the same - a value of 20 kg•cm/s (or cN•s). Since the same spring is used, the same impulse is delivered. Thus, each cart encounters the same momentum change in every situation - a value of 20 kg•cm/s. For the same momentum change, an object with twice the mass will encounter one-half the velocity change. And an object with four times the mass will encounter one-fourth the velocity change.
p7Solving Explosion Momentum Problems
Since total system momentum is conserved in an explosion occurring in an isolated system, momentum principles can be used to make predictions about the resulting velocity of an object. Problem solving for explosion situations is a common part of most high school physics experiences. Consider for instance the following problem:
A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. The cannon is at rest when it is ignited. Immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in 0.0218 seconds. Determine the post-explosion speed of the cannon and of the tennis ball.
Like any problem in physics, this one is best approached by listing the known information.
Given:
||< Cannon: ||< m = 1.27 kg ||< d = 6.1 cm ||< t = 0.0218 s ||
Ball:
m = 56.2 g = 0.0562 kg
The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds. The speed can be assumed constant since the problem states that it was measured after the impulse of the explosion when the acceleration had ceased. Since the cannon was moving at constant speed during this time, the distance/time ratio will provide a post-explosion speed value.
vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded)
The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. If momentum is to be conserved, then the after-explosion momentum of the system must be zero (since the pre-explosion momentum was zero). For this to be true, then the post-explosion momentum of the tennis ball must be equal in magnitude (and opposite in direction) of that of the cannon. That is,
mball • vball = - mcannon • vcannon
The negative sign in the above equation serves the purpose of making the momenta of the two objects opposite in direction. Now values of mass and velocity can be substituted into the above equation to determine the post-explosion velocity of the tennis ball. (Note that a negative velocity has been inserted for the cannon's velocity.)
(0.0562 kg) • vball = - (1.27 kg) • (-280 cm/s)
vball = - (1.27 kg) • (-280 cm/s) / (0.0562 kg)
vball = 6323.26 cm/s
vball = 63.2 m/s
Using momentum explosion, the ball is propelled forward with a speed of 63.2 m/s - that's 141 miles/hour!
It's worth noting that another method of solving for the ball's velocity would be to use a momentum table similar to the one used previously in Lesson 2 for collision problems. In the table, the pre- and post-explosion momentum of the cannon and the tennis ball. This is illustrated below.
MomentumBefore Explosion
MomentumAfter Explosion
Cannon
0
(1.27 kg) • (-280 cm/s)= -355 kg•cm/s
Tennis Ball
0
(0.0562 kg) • v
Total
0
0
The variable v is used for the post-explosion velocity of the tennis ball. Using the table, one would state that the sum of the cannon and the tennis ball's momentum after the explosion must sum to the total system momentum of 0 as listed in the last row of the table. Thus,
-355 kg•cm/s + (0.0562 kg) • v = 0
Solving for v yields 6323 cm/s or 63.2 m/s - consistent with the previous solution method.
Using the table means that you can use the same problem solving strategy for both collisions and explosions. After all, it is the same momentum conservation principle that governs both situations. Whether it is a collision or an explosion, if it occurs in an isolated system, then each object involved encounters the same impulse to cause the same momentum change. The impulse and momentum change on each object are equal in magnitude and opposite in direction. Thus, the total system momentum is conserved.

The importance of rebounding is critical to the outcome of automobile accidents. In an automobile accident, two cars can either collide and bounce off each other or collide, crumple up and travel together with the same speed after the collision. But which would be more damaging to the occupants of the automobiles - the rebounding of the cars or the crumpling up of the cars? Contrary to popular opinion, the crumpling up of cars is the safest type of automobile collision.
Lesson 2d (3/8)
this law.
Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.
The process of solving this problem involved using a conceptual understanding of the equation for momentum (p=m*v). This equation becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity.
Mbefore*Vbefore=Mafter*Vafter
Lesson 2e (3/8)

Force and time are inversely proportional; for the same mass and velocity change. Mass and force are directly proportional Mass and velocity change are inversely proportional.
Lesson 1c (3/6)
velocity change.
First
First we will a collision.
Observe
Observe that the be decreased.
There
There are several a collision.
The same principle explains why dashboards are padded. If the air bags do not deploy (or are not installed in a car), then the driver and passengers run the risk of stopping their momentum by means of a collision with the windshield or the dashboard.
as rebounding.
From
From the impulse-momentum large impulse.
The
The importance of automobile collision.
Lesson 2d (3/8)
The momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in an isolated system, there are no exceptions to this law.
Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.
Lesson 2e (3/8)

Lesson 1b (3/6)
The more momentum that an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed.
p3 IfIf the force object up.
p4These
These concepts are
In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum
collisionsp6OneOne focus of impulse-momentum change equation. Inequation.In a collision,
Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force.
{http://www.physicsclassroom.com/Class/momentum/u4l1a5.gif} Ftp11ForceForce and time
Lesson 1c (3/6)
There are four physical quantities mentioned in the above statement - force, time, mass, and velocity change.
First we will examine the importance of the collision time in affecting the amount of force that an object experiences during a collision.
Observe that the greater the time over which the collision occurs, the smaller the force acting upon the object. Thus, to minimize the effect of the force on an object involved in a collision, the time must be increased. And to maximize the effect of the force on an object involved in a collision, the time must be decreased.
There are several real-world applications of these phenomena. One example is the use of air bags in automobiles. Air bags are used in automobiles because they are able to minimize the effect of the force on an object involved in a collision.
The same principle explains why dashboards are padded. If the air bags do not deploy (or are not installed in a car), then the driver and passengers run the risk of stopping their momentum by means of a collision with the windshield or the dashboard.
Occasionally when objects collide, they bounce off each other as opposed to sticking to each other and traveling with the same speed after the collision. Bouncing off each other is known as rebounding.
From the impulse-momentum change theorem, we could deduce that a rebounding situation must also be accompanied by a large impulse.
The importance of rebounding is critical to the outcome of automobile accidents. In an automobile accident, two cars can either collide and bounce off each other or collide, crumple up and travel together with the same speed after the collision. But which would be more damaging to the occupants of the automobiles - the rebounding of the cars or the crumpling up of the cars? Contrary to popular opinion, the crumpling up of cars is the safest type of automobile collision.
Lesson 2d (3/8)
Lesson 2e (3/8)

Homework
Lesson 1a (3/5)
and velocity.
Momentum = mass • velocity
In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as p = m • v
A doubling of the mass results in a doubling of the momentum.Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in aquadrupling of the momentum.
All questions were answered successfully and efficiently.
Lesson 1b (3/6)
The more momentum that an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed.
p3 If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up.
p4These concepts are merely an outgrowth of Newton's Second Law as discussed in an earlier unit. Newton's second law (Fnet = m • a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. Ft=mΔ V
In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum
collisionsp6One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first law that we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v.
Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force.
{http://www.physicsclassroom.com/Class/momentum/u4l1a5.gif} Ftp11Force and time are inversely proportional; for the same mass and velocity change. Mass and force are directly proportional Mass and velocity change are inversely proportional.
Lesson 1c (3/6)
Lesson 2d (3/8)
Lesson 2e (3/8)

Homework
Lesson 1a (3/5)
and velocity.
Momentum = mass • velocity
p3InIn physics, the rewritten as
p p = m
where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.
p4 TheThe standard metric of momentum.
p5Momentum
Momentum is a vector quantity.
p6From
From the definition an object.
A doubling of the mass results in a doubling of the momentum.Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in aquadrupling of the momentum.
All questions were answered successfully and efficiently.

Homework
Lesson 1a (3/5)
All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast thestuff is moving. Momentum depends upon the variables mass and velocity.
Momentum = mass • velocity
p3In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.
p4 The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units that are acceptable (though not conventional) units of momentum.
p5Momentum is a vector quantity.
p6From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object.
A doubling of the mass results in a doubling of the momentum.Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in aquadrupling of the momentum.
All questions were answered successfully and efficiently.

Homework
Lesson 1a (3/5)

The spring of a dart gun exerts a force on a dart as it is launched from an initial rest position.
PE --> KE
Energy Analysis
Objective
Estimate how much electrical energy you consume on a daily basis (ie: what does it cost my parents)
5 kWh
What to do?
Identify 10 electrical appliances you use on a daily basis and identify their power rating (in Watts).
{Screen_shot_2012-02-09_at_10.04.28_AM.png}
Apple laptop charger= 60 W= .06 kW
Flat Iron= 25 W= .025 kW
iHome= 150 W= .150 kW
iPhone Charger= 10W =.10 kW
Blow Dryer= 1875 W= 1.875 kW
Vizio Tv= 110 W=.110 kW
Lamp= 60 W= .06 kW
Microwave= 950 W=.95 kW
Pencil Sharpener= 20 W= .02 kW
Apple Desktop Computer= 120W= .120kW
The power rating will be listed on the info label – typically on the back of or under the appliance. Often the power consumption is listed in terms of AMPERAGE (amperage measures the current of electricity consumed). The power rating (in Watts) may be calculated using the following formula:
Power (in Watts) = AMPS x 120 volts.
For example, a 12 AMP vacuum cleaner uses 12A x 120v = 1440 Watts of power.
Determine the total kWh (kilowatt-hours) of power that you consume with these ten appliances.
First estimate how many hours you use each appliance. For example, you may use your hairdryer every second day for 10 minutes – so on average you use it for 0.17h/d ¸ 2 = 0.085 h per day.
12h/d
.167h/d
2 h/d
6h/d
3h/d
3h/d
.33h/d
.167h/d
.0167h/d
12h/d
Calculate the kWh used per day by multiplying the wattage of the appliance by the hours used (per day).
{Screen_shot_2012-02-09_at_10.05.00_AM.png}
.72 kWh
.0042 kWh
.30 kWh
.60 kWh
5.625 kWh
.330 kWh
.0198 kWh
.15865 kWh
.000334 kWh
1.44 kWh
Finally, determine the total kWh you use per day.
9.197984 kWh
What does this cost your parents?
$0.119164*9.197984= $1.10
Look at your household electric bill. Find two numbers:
the total dollar amount due
$265.38
kWh consumed that month
1424 kWh
Costs of Electricity.
Calculate the average cost per kWh by dividing the dollar amount by the kWh consumed. This is approximately what each kWh costs your family. (Note: this figure includes fees and taxes, too). The cost will probably be between $0.10-$0.20 per kWh – depending on how much electricity you use. Electricity costs more if you exceed a certain usage.
$0.186362
You can also look this value up on the bill itself.
$0.119164
Compare the two values. Why are they different?
Probably because PSE&G wants to make a profit.
Multiply the kWh you use per day by the cost per kWh to determine what you cost your parents.
9.197984* $0.119164= $1.10
Evaluate
Does the cost of electricity seem like a lot to you? Discuss this with your friends.
It doesn’t seem that much but in the long run it probably adds up. There are so many appliances that are left out so I’m sure that it is much more.
Keep in mind that this cost only reflects ten appliances. What do you think your total cost actually is? Are there major appliances that you may have excluded from your list of ten that would make a big difference?
I left out refrigerator, garage door, pool heater, etc. I am sure it costs much more than this. This definitely would have made a huge difference.
If you think your consumption is a lot – what can you change? Do you need/should change?
I should probably unplug my chargers and such. I always leave everything plugged in—it’s a very bad habit. However, I have gotten better at shutting off the lights.
I think that the electric bill can be a little overpriced. I think that it especially hard for those who have big families, who are constantly maximizing the amount of electricity used. Also, when children are younger, I feel like there are more lights on in the house and more use of electrical appliances (tv, video games). Ultimately, I feel like people underestimate and overlook how much each appliance costs, and are overall uneducated.