Compute for the total mechanical energy of a 3.5 kg baseball dropped from a height of 3 meters moving at a speed of 10 m/s. NEED ASAP GRADE 8 SCIENCE TY IN ADVANCE :))

Answers

Answer 1

Mechanical energy = Kinetic Energy + Potential Energy

Kinetic Energy :

Kinetic energy = 1/2 * m * v² = 1/2 * 3.5 * (10)²

=> Kinetic energy = 175 J

Potential Energy :

Potential energy = m * g * h = 3.5 * 10 * 3 = 105 J

Hence,

Mechanical energy = Kinetic Energy + Potential Energy

= 175 J + 105 J

= 280 J

Answer 2

Taking into account the potencial, kinetic and mechanical energy, the total mechanical energy is 278.005 J.

Potencial energy

Potential energy refers to the position that a mass occupies in space and its magnitude is directly proportional to the height at which the object is, with respect to a reference position, to the mass of the object and to gravity:

Ep = m×g×h

Kinetic energy

On the other hand, kinetic energy manifests itself when bodies move and is associated with speed. The kinetic energy of a body depends on its mass and the square of its speed:

Ec = [tex]\frac{1}{2}[/tex]×m×v²

Mechanical energy

The mechanical energy of a body is the capacity it has to carry out mechanical work, that is, to produce a movement due to causes of mechanical origin, such as its position or its speed.

The mechanical energy of a body is the sum of its kinetic energy and its potential energy:

Em = Ep + Ec

Where

Em is the mechanical energy (J).Ep the potential energy (J). Ec the kinetic energy (J).

Then:

Em= m×g×h + [tex]\frac{1}{2}[/tex]×m×v²

Principle of conservation of mechanical energy

The Principle of conservation of mechanical energy indicates that, in the absence of friction and without the intervention of any external work, the sum of the kinetic and potential energies remains constant, that is, the mechanical energy of a body remains constant throughout the movement. if no friction force acts.

This case

In this case, you know that:

m= 3.5 kgg= 9.81 [tex]\frac{m}{s^{2} }[/tex]h= 3 mv= 10 [tex]\frac{m}{s}[/tex]

Complying with the Principle of Conservation of Mechanical Energy and replacing in the expression of mechanical energy:

Em= 3.5 kg×9.81 [tex]\frac{m}{s^{2} }[/tex]×3 m + [tex]\frac{1}{2}[/tex]×3.5 kg×(10 [tex]\frac{m}{s}[/tex])²

Solving:

Em=103.005 J + 175 J

Em= 278.005 J

Finally, the total mechanical energy is 278.005 J.

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Answer:

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When conducting a polygraph test why is it important for the interviewer to establish a baseline on the interviewee?

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Answer:

Because always when we do an experiment, we need something to compare.

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Most of the Earth's volcanoes occur
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near the center of tectonic plates.
B.
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D.
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Answer:

A.

Explanation:

NEAR THE CENTER OF TECTONIC PLATES.

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Explanation:

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accelerates uniformly from rest, reaching a speed of 36 meters per second in 6.0 seconds

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Answer:

What's the question?

Explanation:

When does The mass of an object change
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V=A shot-putter tosses an 8 kg metal ball across a field. The ball moves from rest to a velocity of 4 m/s in 0.8 seconds. Calculate the force (in Newtons) that was used to move the ball. Don't input the unit.

Answers

This one is a little tricky but ima see if I can help

This is a short answer question, please answer in full sentences.
Apply Newton’s third law of motion to the action of serving a volleyball.

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Answer:

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A sound wave ________ and ______ the air molecules around your ear.

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A sound wave squeezes together and thins out the air molecules around your ear.  The places where these things happen are called the compressions and rarefactions of the sound wave.

A car moves with a velocity of 20 m/s to east and the other with velocity of 15 m/s to west. If they move from the same point, how far will each of them travel in 2 minutes? Calculate the distance between them after 2 minutes​

Answers

Answer:

4.2 km

Explanation:

CAR A(velocity =20m/s to East)

In 1s, it covers 20m.

In 120s(2min),it covers20*120m.=2400m=2.4 km

CAR B(velocity =15m/s to west)

In 1s, it covers 15m.

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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (a) Show that the frictional torque in the system is about 5 kN m. (3) 1. (b) When the power is switched off, the roundabout decelerates uniformly because the

Answers

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

What is the speed of a 55.0 kg skydiver who has 7.81 x 104 J of kinetic energy?

Answers

Answer:

v=53.3m/s

Explanation:

Ek=1/2mv²

7.81×10⁴=1/2×55.0v²

v= the square root of 7.81×10⁴/0.5×55.0

v=53.3m/s

The speed of a 55.0 kg skydiver who has 7.81 x 104 J of kinetic energy is 53.29 m/s.

To find the speed, the given values are,

Mass m = 55 Kg

Kinetic Energy = 7.81x 10⁴ J

What is Speed?

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

 Speed  = Distance / Time m/s.

The unit is m/s or km/hr or mile/hr.

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Formula,

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When we solve for v,

v² = 2Ke/m

Substituting the given values,

v² = 2(7.81 x 10⁴) / 55

Simplifying,

v² = 15.62 x 10⁴ / 55

v² = 0.284 x 10⁴

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v = 53.29 m/s.

The speed of a 55.0 kg skydiver who has 7.81 x 104 J of kinetic energy is 53.29 m/s.

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Question 10 of 10
Two particles are separated by 0.38 m and have charges of 1.25 x 10-9C and
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Explanation:

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where F = force and S = displacement

And q = angle between F and S.

Hope this helps! <3

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Newton (N)

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Answer:

b ok bye thansks

bye.

Always point the open end of a test tube being heated d. Away from all people.

When heating a substance in a lab, there is a chance that the substance could react to the heat and evaporate into a gaseous state.

This could be very dangerous if the gas is:

corrosivea pollutant biological breathing hazard

You should therefore be careful to point the tube away from others and yourself so that the gas does not affect either them or you.

In conclusion therefore, always point the open end of the tube away from all people.

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Answer:

1 Imadge 2 no

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Answer:

atoms

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.....(✿^‿^).......

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from problem 2,suppose another boy, boy C pulls the heavy cabinet with 5N of force in the same direction with boy A

1. what will be the net foece on the cabinet?
2. will the cabinet move?
3. in what direction will the cabinet move?​

Answers

Answer:

Searching in Google I found that in problem 2 were two boys pulling the heavy cabinet, a Boy A pulling in the left direction with 10 N of force, and a boy B pulling in the right direction with 5 N.  

1. F = 10 N

2. Yes, it will move.

3.  It will move to the left direction of the cabinet (same direction as the boys A and C).

           

Explanation:

Searching in Google I found that in problem 2 were two boys pulling the heavy cabinet, a Boy A pulling in the left direction with 10 N of force, and a boy B pulling in the right direction with 5 N.  

1. Hence, if another boy (C) pull in the same direction as the boy A with 5 N of force, we have:

[tex] \Sigma F = F_{A} + F_{C} - F_{B} [/tex]

[tex] \Sigma F = 10 N + 5 N - 5 N = 10 N [/tex]

The net force on the cabinet is 10 N.

2. Yes, the cabinet will move. From point 1 we can deduce that the cabinet will move since the net force on the cabinet is different from zero (10 N).                    

3. The direction of the motion is the same as the Boys A and C, that is to say, to the left of the cabinet.      

I hope it helps you!

The net force acting on an object, is the vector sum of all the forces acting on it

The correct values for the forces acting on the cabinet are;

1. The net force on the cabinet is 10 N

2. Yes, the cabinet will move

3. The cabinet will move in the direction of boy A and boy C

The reason the above values are correct are as follows:

Question: The possible missing part of the question is; In problem 2 boy A is pulling a heavy cabinet with [tex]F_A[/tex] = 10 N force, and boy B is pulling the same cabinet at the same time and in opposite direction with a force of [tex]F_B[/tex] = 5 N

The known parameter:

The force with which boy C pulls the heavy cabinet = 5 N

The direction in which boy C pulls the cabinet = The same direction as boy A

Question 1. The net force on the cabinet, [tex]F_{NET}[/tex] is given as follows;

Taking the direction of the force of boy A as the positive direction, we have;

[tex]\overset \longrightarrow {F_A}[/tex] = 10 N

[tex]\overset \longrightarrow {F_B}[/tex] = - 5 N

[tex]\overset \longrightarrow {F_C}[/tex] = 5 N

[tex]\overset \longrightarrow {F_{NET}}[/tex] = [tex]\overset \longrightarrow {F_A}[/tex] + [tex]\overset \longrightarrow {F_B}[/tex] + [tex]\overset \longrightarrow {F_C}[/tex]

[tex]\overset \longrightarrow {F_{NET}}[/tex] = 10 N + (-5 N) + 5 N = 10 N

The net force on the cabinet, [tex]\overset \longrightarrow {F_{NET}}[/tex] = 10 N

Question 2. Yes: Given that there is a net force acting on the cabinet, according to Newton's first law of motion, in the absence of friction force, the cabinet will move in the direction of the net force of 10 Newton

Question 3. The cabinet will move in the direction of the net fore

The direction of the net force is the direction boy A and boy C are pulling the cabinet

Therefore, the cabinet will move in the direction boy A and boy C are pulling the heavy cabinet

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2. Floyd Beattie set an unofficial speed record for a Floyd Beattie set an unofficial speed record for a unicycle in 1986. He rode the unicycle through a 200
m speed trap and accelerated at a constant rate along the speed trap, so that his initial speed was 9.78
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Answers

Answer:

0.022m/s²

Explanation:

Given

v = 10.22m/s

u = 9.78m/s

S = 200m

Using the equation of motion'

v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

S is the displacement

Substitute the given parameters into the formula:

10.22² = 9.78² + 2a(200)

104.4484 = 95.6484 + 400a

400a = 104.4484 -95.6484

400a = 8.8

a = 8.8/400

a = 0.022m/s²

Hence the magnitude of his acceleration would have been 0.022m/s²

How is a warm front different than a cold front? ANSWER NOW I ONLY HAVE 15 MINUTES LEFT! I will not give brainliest unless answered in time

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Answer:

Warm fronts move more slowly and are less violent than cold fronts. They are associated with warm air moving over cold air and are more likely to produce large regions of light to moderate rain, drizzle or snow.

Explanation:

HELP PLEASE!!!! 20 POINTS! If we increase the distance traveled over the same period of time, this will (2 points) decrease the speed
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Increase the speed of it

A train traveled at an average speed of 80 mph for 4.2 hours. Then, it traveled at an average speed of 60 mph for 1.5 hours. What was the total distance covered by the train?

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Answer:

The answer is four hundred twenty six or 426

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Acceleration is a vector quantity. When an object is said to have a negative acceleration, does it always mean that the object is slowing down? OTrue OFalse​

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The answer to that would be False.

.
What change in the velocity's horizontal
component (v) and the range occurs
with each unit of time while getting
through the water?

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Answer:

h the books

Explanation:

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Scientists might make a computer model of volcanic eruptions. What is the biggest benefit of this model

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Answer:

I think it would be to be able to prevent a volcanic eruption in the future and take the appropriate prevention protocols.

Explanation:

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