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Helen [10]
2 years ago
7

Tasks

Physics
1 answer:
barxatty [35]2 years ago
7 0

Answer:

NE DIYON INGILIZ MISIN SEN

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How do I do this physics problem about potential energy and kinetic energy?
larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

4 0
3 years ago
What is the equation for an inelastic collision
abruzzese [7]
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
7 0
3 years ago
The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia
MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

              How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

                    = 1 J - 0.313 J

                    = 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

                    = 1 J - 0.45 J

                    = 0.55 J

So,

Energy Lost for group B's car = 0.55 J

8 0
3 years ago
I WILL GIVE BRAINLIEST
11111nata11111 [884]

Answer:

A, B, F

Explanation:

I believe these are the answers, sorry if it is incorrect.

3 0
3 years ago
Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d
Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

8 0
3 years ago
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