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marissa [1.9K]
2 years ago
7

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

determine the distance from the edge of the wall to the water's surface.
Physics
2 answers:
Effectus [21]2 years ago
7 0
Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer:  11.025 m
yawa3891 [41]2 years ago
4 0

Answer: 11.025

Explanation:

Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.

Because the pebble hits the surface of the water after 1.5 s, therefore

h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

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Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed v_i = 42 m/s

final speed v_f = 0

time taken by the car to stop

t = 4s

now the distance moved by the car before it stop is given as

d = \frac{v_f + v_i}{2} \times t

now we have

d = \frac{42 + 0}{2} \times 4

d = 84 m

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from
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Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

m - Mass of the cat, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

If we know that m = 2.70\,kg, g = 9.807\,\frac{m}{s^{2}}, h = 5.20\,m and W_{l} = 120\,J, then the final speed of the cat is:

v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }

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The speed of the cat when it hits the ground is approximately 7.586 meters per second.

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4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes
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Answer:

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The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis
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Answer:

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