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2 years ago

Using the law of conservation of energy, what will be the KE of an arrow having a PE of 65J after it is shot from a bow?

Physics

1 answer:

Colt1911 [192]2 years ago

7 0

Answer:

KE = 65 J (The potential energy stored in the stretched bow is converted to kinetic energy of the moving bow)

Explanation:

Law of conservation of energy states that energy can neither be created nor be destroyed but transferred from one form to another. In other words, the total energy of the system at any point is always a constant.

Now, when the bow is in stretched position, the energy associated with the stretched bow is potential energy (PE).

The energy associated with motion of the body is kinetic energy (KE).

So, the stretched bow is not moving. So, KE at the start is 0.

Therefore, total energy initially is given as:

Initial total energy = PE + KE = PE + 0 = 65 J

Now, when the bow is shot, the bow is in motion. So, the total energy of the bow is only due to kinetic energy associated with it now. Also, the potential energy of a moving bow is 0.

Therefore, final total energy is given as:

Final total energy = Final KE + Final PE = Final KE + 0 = Final KE

Now, we know, from conservation of energy,

Initial total energy = Final total energy

⇒ 65 J = KE

or Final KE = 65 J

Therefore, the KE of an arrow having a PE of 65J after it is shot from a bow is 65 J.

Hence, the potential energy stored in the stretched bow is converted to kinetic energy of the moving bow.

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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle $F_{net}$ is given by $F_{net}=F-f$

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for $F_{net}$

$F_{net}=ma$ hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

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3 years ago

Snow and sleet when they fall to the ground is solar energy or gravitational force?

Nikolay [14]

Gravitational energy

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2 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

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3 years ago

Indicate whether each of the following statements is true or false. In a coal-fired power plant, chemical energy is first conver

denis23 [38]

Answer:

1. In a coal-fired power plant, chemical energy is first converted to thermal energy. TRUE.

The chemical energy in the coal is converted to thermal energy when the coal is burnt to produce steam.

2. In a coal-fired power plant, approximately 2/5 of original energy in the coal is lost to heat. FALSE.

Approximately 3/5 of the original energy is lost not 2/5 so this statement is false.

3. The molecules in cold air move faster than in hot air. FALSE.

Molecules with more heat move faster than molecules with less. Molecules in cold air therefore, will move slower than molecules in hot air.

6 0

2 years ago

Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti

Gwar [14]

After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.

==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
(1,500) · (0.966)^x
lumens of light remaining.

=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".

What does that mean ? Break it down: 15dB in 100 meters is 0.15dB per meter.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.

Look at this ==> 10 log(0.966) = -0.15  <== loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...

-- 0.15 dB loss per meter

-- 'x' meters of cable

-- 0.15x dB of loss.

If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .

and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = 8.3% <== That's how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.

5 0

2 years ago