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3 years ago

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Physics

1 answer:

Sladkaya [172]3 years ago

3 0

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

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ivann1987 [24]

Answer:

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Explanation:

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

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3 years ago

What is the role of equations in this course?

Karolina [17]

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3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

Charge of an Electron

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

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Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

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Summarizing, the following are the possible values for the charge of an oil droplet:

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3 years ago

A student drops an object out the window of the top floor of a high-rise dormitory.

seropon [69]

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Now we can switch over to the metric system, where the acceleration
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D = (1/2) (9.8) x (4)² = 78.4 meters

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