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muminat
3 years ago
9

Suppose that instead of dropping the rock you throw it downwards so that its speed after falling 7 meters is 23.43 m/s. How much

kinetic energy did you give the rock in your throw

Physics
2 answers:
Mkey [24]3 years ago
8 0

Answer:

205.88 ×mass of rock

This problem uses the expression for the kinetic energy of a moving body. K.E = 1/2mv², where m = mass of rock and v = velocity of the rock.

Explanation:

We are requested to calculate the kinetic energy but in order to do sk we must first calculate the initial velocity at which the rock was thrown downwards.

Given the following: h = -7m (thrown downwards and assuming upwards to be positive), v = -23.43 (same reason) and g = 9.8m/s².

The full calculation can be found in the attachment below.

Shkiper50 [21]3 years ago
5 0

Answer:

3,298.1 Hz

Explanation:

Solution is attached

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A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee
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Answer:

θ = sin⁻¹\sqrt{2gd}

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = \sqrt{2gd}

θ = sin⁻¹ \sqrt{2gd}

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3 years ago
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Both magnitude and DIRECTION
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3 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0
3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

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c) gravitational force acting

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4 0
3 years ago
A mover loads a 100 kg box into the back of a moving truck by
NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

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Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

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Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

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MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

3 0
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