We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.

Answers

Answer 1

Answer:

Answer:

* roller skates and ice skates.

* roller coaster

Explanation:

One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.

Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.

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The energy provided each hour by heat to the turbine in an electric power plant is 9.0×10^12 J. If 5.4 × 10^12 J of energy is exhausted each hour from the engine as heat, what is the efficiency of this heat engine?

Answers

Answer:

60 %

Explanation:

Efficiency is defined as the ratio of output power to the input power.

Input energy each hour = 9 x 10^12 J

Output energy each hour = 5.4 x 10^12 J

Efficiency = Output energy per hour / input energy per hour

Efficiency = (5.4 x 10^12) / ( 9 x 10^12) = 5.4 / 9 = 0.6

Efficiency in percentage form = 0.6 x 100 = 60 %

Final answer:

The efficiency of a heat engine is calculated using the formula (Energy Input - Energy Output) / Energy Input. Given the figures, the efficiency of the engine is 40%, indicating that 40% of the input energy is converted into work.

Explanation:

The efficiency of a heat engine is determined by the ratio of work output to energy input. In the given scenario, the turbine in an electric power plant is supplied with 9.0 x 10^12 joules of energy, and 5.4 x 10^12 joules of energy is expelled as heat per hour. We can calculate the efficiency using the equation:

Efficiency = (Energy Input - Energy Output) / Energy Input

By substituting the given values, Efficiency = (9.0 x 10^12 J - 5.4 x 10^12 J) / 9.0 x 10^12 J = 0.4 or 40%

This means the heat engine of the power plant has a 40% efficiency, meaning 40% of the energy input is converted into work while 60% is discarded as waste heat.

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A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

A 84-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.73 as. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and then comes to rest.What does the spring scale register During the first 0.80s of the elevator’s ascent?

Answers

Answer:

$SR=949.2N$

Explanation:

From the question we are told that:

Mass $M=84kg$

Speed $V=1.2m/s$

Acceleration Time $t_a=0.73$

Constant speed Time $t_s=5.0s$

Deceleration time $t_d=1.4s$

Generally the equation for Acceleration is mathematically given by

$a=(v)/(t)$

Therefore acceleration for the first 0.80 sec is

$a=(1.2)/(0.80)$

$a=1.5m/s^2$

Therefore

Spring Reading=Normal force -Reaction

$SR=m(g+a)$

$SR=84(9.8+1.5)$

$SR=949.2N$

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?

Answers

The lightning bolt that moves 50.0 C of charge is converted to qe units. The conversion factor is 1 coulumb is equivalent to 6.24150975·10^18 e. Thus, converting 50 C to e is:

50.0 C * (6.24150975x10^18 e)/1C
equals 3.120754875x10^20 e

Answer: There are $3.125\tims 10^(20)$ number of electrons.

Explanation:

We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

$nq_e=Q$

where,

n = number of electrons

$q_e$ Charge of one electron = $1.6* 10^(-19)C$

Q = Total charge = 50 C.

Putting values in above equation, we get:

$n* 1.6* 10^(-19)C=50C\n\nn=(50C)/(1.6* 10^(-19)C)\n\nn=3.125* 10^(20)C$

Hence, there are $3.125\tims 10^(20)$ number of electrons.

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answers

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

$\alpha =atan((0.5)/(1))=26.56°$

$D=√(1^2+0.5^2)=1.118km$

The realitve velocity of the boat respect to the water is:

$V_(B/W)=[3*cos\beta ,3*sin\beta ]$ where β is the angle it has to be pointed at.

From the relative mvement equations:

$V_(B/W)=V_B-V_W$ where $V_B=[V*cos\alpha ,-V*sin\alpha ]$

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

$(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2$

$V^2-4*V*sin\alpha -5=0$ Solving for V:

V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:

$\beta =acos((V*cos\alpha )/(3) ) = 10.29°$

The amount of time:

$t = D/V = (1118m)/(3.3m/s) =338.7s$

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