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Flura [38]
3 years ago
13

To test the resilience of its bumper during low-speed collisions, a 4 060-kg automobile is driven into a brick wall. the car's b

umper behaves like a spring with a force constant 8.00 106 n/m and compresses 3.72 cm as the car is brought to rest. what was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Simora [160]3 years ago
4 0
Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
Spring compression, x = 3.72 cm = 3.72 x 10⁻² m

Let the car strike the wall with speed v m/s.

The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
(1/2)mv² = (1/2)kx²
(4060 kg)*(v m/s)² = (8 x 10⁶ N/m)(3.72 x 10⁻² m)²
4060 v² = 1.1701 x 10⁴
v = 1.6513 m/s

Answer:  1.65 m/s (nearest hundredth)

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The gravitational attraction between two objects increases if
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3 years ago
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.65 m/s. The stone subs
xxTIMURxx [149]

Answer:

1. 20.54m/s

2. 1.52s

Explanation:

QUESTION 1:

The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:

v² = u² + 2as

Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

v² = 5.65² + 2 (9.8 × 19.9)

v² = 31.9225 + 2 (195.02)

v² = 31.9225 + 390.04

v² = 421.9625

v = √421.9625

v = 20.5417

v = 20.54m/s

QUESTION 2:

Using v = u + at

Where v = final velocity (m/s) = 20.54m/s

t = time (s)

u = initial velocity (m/s) = 5.65m/s

a = acceleration due to gravity (m/s²)

v = u + at

20.54 = 5.65 + 9.8t

20.54 - 5.65 = 9.8t

14.89 = 9.8t

t = 14.89/9.8

t = 1.519

t = 1.52s

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