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3 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]3 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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Which example best represents translational kenetic energy

Mila [183]

Answer:

an apple falling off a tree

Explanation:

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3 years ago

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

3 years ago

Please help I need this fast

STALIN [3.7K]

Answer:

0 m/sec

Explanation:

b/c they were at rest and initial means at rest ,at rest means 0 HOPE THIS HELPS

8 0

3 years ago

Read 2 more answers

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

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3 years ago

While looking at a cliff, you observe that three visible layers of rocks are tilted about 30 degrees. There are four straight ho

xxTIMURxx [149]

Answer:

the principle of original horizontality and the principle of superposition

Explanation:

The principle of horizontality states that layers of sediment are originally deposited horizontally under the influence of gravity.

The principle of superposition states that the oldest layer layer is at the bottom and each layer above it is younger, with the youngest being at the top.

Unconformities help us find the age of different layers. An unconformity is a surface in which no new solid matter is deposited after a long geologic interval. Angular unconformity is a type of unconformity which different kinds of stratum were tilted or folded before deposition of younger layers of solid matter above the unconformity. Once the layers were folded and tilted, the older layers of the solid matter eroded, then the younger layers were deposited on the older layers. There angular unconformity is the contact between young and old layers of solid matter.

Therefore, these two principles therefore describe how the tilted layers are older than horizontal layers.

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3 years ago