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dlinn [17]
3 years ago
8

If you dropped a ball off the roof, and could neglect air resistance, the impact velocity of the ball on the sidewalk would be 1

02 miles per hour [mph]. Calculate the height of the building in units of feet.
Physics
1 answer:
Orlov [11]3 years ago
6 0

Answer:

h = 105.98 m

Explanation:

We will use the third equation of motion here:

2gh= v_f^2-v_i^2

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = ?

vf = final speed = (102 mi/h)(1 h/3600 s)(1609.34 m/1 mi) = 45.6 m/s

vi = initial speed = 0 m/s

Therefore,

2(9.81\ m/s^2)h = (45.6\ m/s)^2-(0\ m/s)^2\\\\h = \frac{ (45.6\ m/s)^2}{2(9.81\ m/s^2)}

<u>h = 105.98 m </u>

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A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles
slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

(c) The kinetic energy when its position is 2.50 cm is 0.0195 J

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035  × v² + 1/2 ×  40 × 0.012²

0.032 - 1/2 ×  40 × 0.012² = 1/2 × 0.035  × v²

0.02912 = 1/2 × 0.035  × v²

1/2 × 0.035  × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm,  v ≈ 1.28996 m/s

(c) When its position is 2.50 cm = 0.025 m, we have;

E = PE + KE

0.032 = 1/2 ×  40 × 0.025² + KE

KE = 0.032 - 1/2 ×  40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

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Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
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