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Aleks04 [339]
4 years ago
9

If I have a picosecond laser how would that be expressed in terms of zeptoseconds? In terms of petaseconds?

Physics
1 answer:
kolbaska11 [484]4 years ago
3 0
Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds  =  10^15  seconds

To express Picoseconds into any of other two, you have to divide  10^-12 by the power index of the one in question

1Picosecond :   10^-12  /  10^-18 =    10^ (-12- 18) =  10^ (-12+18)= 10^6  zeptoseconds

1Picosecond :   10^-12  /  10^15 =    10^ (-12-15) =  10^-27  Petaseconds.

1Picosecond = 10^6  zeptoseconds

1Picosecond = 10^-27  Petaseconds




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Form a hypothesis. Which will be greater, the potential energy of the car at the top of the ramp or the kinetic energy of the ca
AleksandrR [38]

The potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp

Explanation:

Here we have a car moving down along a ramp. In absence of force of friction, the total mechanical energy of the car while moving down would be conserved:

E=PE+KE = const.

where

PE is the potential energy

KE is the kinetic energy

However, this is not true if friction acts on the car. When the car is still at the top of the ramp, its speed is zero, so its kinetic energy is zero, and all the energy is just potential energy:

E=PE

While the car moves down along the ramp, friction does work on it, and part of the total mechanical energy is wasted and converted into thermal energy. As a result, the car reaches the bottom of the ramp with a final energy which is less than the initial energy:

E'

Also, at the bottom of the ramp, all the energy is just kinetic energy, since the height of the car is now zero, so

E'=KE

It follows that

KE

So, the potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp.

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6 0
3 years ago
Friction occurs when the and of two surfaces stick to each other
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Answer:

yes, kinda like when babies are made, there is friction

Explanation:

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3 years ago
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D. What is the net force on the bowling ball rolling lane
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Answer:

Friction.

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3 years ago
A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan
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7 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
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