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Mars2501 [29]
4 years ago
15

Which describes the difference between a bowling ball sitting in the rack waiting to be used and the bowling ball knocking pins

down? A. In the rack, the bowling ball has only potential energy. As it knocks pins down, its potential energy has decreased, while its kinetic energy has increased.
B. In the rack and when knocking down pins, the potential energy and the kinetic energy of the bowling ball are equal.

C. In the rack, the bowling ball has no energy at all. As it knocks down the pins the energy is 75% kinetic and 25% potential.

D. In the rack, the bowling ball has only kinetic energy. As it knocks pins down, its potential energy has increased, while its kinetic energy has decreased.
Physics
2 answers:
777dan777 [17]4 years ago
7 0
The answer is A. In the rack, the bowling ball has only potential energy. As it knocks pins down, its potential energy has decreased, while its kinetic energy has increased.
Alexeev081 [22]4 years ago
6 0
A. In the rack, the bowling ball has only potential energy. As it knocks pins down, its potential energy has decreased, while its kinetic energy has increased.
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you will For this problem, you will need to look up physical parameters for objects in space want to keep about 4 significant fi
Tanzania [10]

Answer:

a)  r₁ = 3.836 10⁷ m,  b)   F = - 3,390 10⁸ N , c)  R = 120.3 m

Explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

       F₁ -F₂ = 0

       F₁ = F₂

       G m M_{m} / r₁² = G m M_{e} / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

         r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

        M_{m} / r₁² = M_{e} / (D - r₁)²

       (D² - 2 D r₁ + r₁²) = M_{e} /M_{m} r₁²

       (1 - M_{e} / M_{m})r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

       (1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

       -80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

        r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

        r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

        r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

       r₁’= 38.355 10 6 m

       r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is    r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

        R = 2 3.8355 10⁷ = 7.671 10⁷ m

        F = F₁ - F₂

        F = G m  M_{m} / R² - G m  M_{e} / (D- R)²

        F = G m ( M_{m} / R² -  M_{e} / (D-R)²)

   F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

        F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

        F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

        F = - 3,390 10⁸ N

c) let's use gravitational force from the moon

        F = G m  M_{m} / R²

        R =√ G m  M_{m} / F

        R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

        R = √ (1.4482 10⁴)

        R = 1.2034 10² m

        R = 120.3 m

8 0
3 years ago
A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it
Sergeeva-Olga [200]

Answer:\theta =45.73^{\circ}

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

T\cos\theta =mg-----1

T\sin \theta =\frac{mv^2}{r}------2

Divide 2 & 1

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{4.65^2}{2.15\times 9.8}

tan\theta =1.026

\theta =45.73^{\circ}

5 0
3 years ago
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