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Ann [662]
3 years ago
14

The initial kinetic energy imparted to a 0.97 kg bullet is 1150 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

sistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained. Answer in units of km.
Physics
1 answer:
lys-0071 [83]3 years ago
6 0

Answer:

113.58 m

Explanation:

Range = Maximum Height

\frac{u^{2}Sin2\theta }{g}=\frac{u^{2}Sin^{2}\theta }{2g}

tan θ = 4

θ = 76 degree

Let u be the velocity of projection

K = 1/2 x m x u^2

1150 = 0.5 x 0.97 x u^2

u = 48.7 m/s

Range = [tex]\frac{u^{2}Sin2\theta }{g}

R = (48.7 x 48.7 x 2 x Sin 76 x Cos 76) / 9.8 = 113.58 m

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(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
 
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                    R = sqrt (400 + 289)
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The direction of the wind will have to be angle between the boat and the resultant.
                   cos θ = (20 knots)/(26.24 knots)
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Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero. 
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3 years ago
3. Two balls are rolling toward each other. One has a momentum of 85kg*m/s, and the other has a momentum of -85kg*m/s. What will
irakobra [83]

Answer:

The total momentum is zero.

Explanation:

This problem can be solved by applying the momentum conservation theorem and the amount of motion. This theorem tells us that the amount of motion is conserved before and after a collision.

In the next equation, we will write to the left of the equal sign the amount of motion before the collision and to the right the amount of motion after the collision.

(P_{1})-(P_{2})=P_{3}

where:

P₁ = momentum of the ball moving to the right, before the collision = 85 [kg*m/s]

P₂ = momentum of the ball moving to the left, before the collision = - 85 [kg*m/s]

P₃ = Final momentum after the collision [kg*m/s]

(85) - 85 = P_{3}\\P_{3}= 0

There is no movement of any of the balls, they remain at rest after the impact.

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2 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
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Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

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b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

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d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

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Answer:

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