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Ne4ueva [31]
1 year ago
12

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
Physics
1 answer:
EleoNora [17]1 year ago
5 0

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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A large bottle contains a number of medicinal tablets is 0.5kg. Calculate the number of tablets in the bottle?
Alexeev081 [22]

Answer:

the answer is 5k in the bottle have

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Is there a definite end to our atmosphere?
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There is no definite end to earths atmosphere, but technically the border between the outer space and earth gets thinner as you move up from the earths surface. The Karman line is the closest definition there is which describes the end of the earth's atmosphere, it is 100 km above earth's sea level at approximately 1.56 % of total earth's radius. This describes the boundary between the outer space and the atmosphere.
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16) A man ran a 5 mile race. The race looped around a city park and back
finlep [7]

Answer:

The man's total displacement is equal to 0.

Explanation:

Given that,

A man ran a 5 mile race. The race looped around a city park and back  to the starting line.

We need to find the total displacement of the man.

We know that,

Displacement = shortest path covered

Also,

Displacement = final position - initial position

As it reaches back to its starting line, it means, the displacement is equal to 0.

Hence, the man's total displacement is equal to 0.

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. The artificial sweetener NutraSweet is a chemical called aspartame (C14H18N2O5). What is (a) its molecular mass (in atomic mas
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The molecular mass, in atomic mass unit, of aspartame would be 294 amu while the mass, in kg, of an aspartame molecule would be 0.294 kg

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(a) Molecular weight = (12x14) + (1x18) + (14x2) + (16x5)

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(b) Mass of 1 molecule of aspartame = mole x molar mass

                                    = 1 x 294

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Converting 294 g to kg = 294/1000

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More on mole and molar mass can be found here: brainly.com/question/6613610

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