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g100num [7]
3 years ago
12

if a stone with an original volcity of 0 is falling from the ledge and takes 8 seconds to hit the ground what is the final volci

ty of the stone
Physics
1 answer:
Nookie1986 [14]3 years ago
7 0
On Earth, gravity adds 9.8 m/s to the downwsrd speed of any falling object, every second. Beginning from zero downward speed, the speed grows to (8 x 9.8) = 78.4 m/s downward after 8 sec.
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Please Help <br><br> Conservation of energy
Sonja [21]

Answer:

a principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

8 0
3 years ago
A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebbl
Leona [35]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have: KE = PE (potential energy) </em>

<em>PE = m x g x h </em>

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\  \frac{m_1}{m_2\times 4} = 1 \\ \\  \frac{m_1}{m_2} = 4

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

8 0
2 years ago
¿Qué es la velocidad?
egoroff_w [7]

Velocidad-es una cantidad vectorial física; Tanto la magnitud como la dirección son necesarias para definirlo.

5 0
3 years ago
A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
3 years ago
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
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