Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
The kinetic energy of the safe increases the force exerted by the concrete
to several times the weight of the safe.
- The magnitude of the force exerted on the safe by the concrete on the is approximately [tex]\underline{29.\overline 3 \, \mathrm{MN}}[/tex]
- The concrete exerts a force that is approximately 1,359.16 times the weight of the safe.
Reasons:
First part
The mass of the steel safe, m = 2,200 kg
Velocity of the safe just before it hits the concrete, v = 40 m/s
The amount by which the safe was compressed, d = 0.06 m
The kinetic energy, K.E., of the safe just before it hits the round is therefore;
[tex]\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}[/tex]
[tex]\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules[/tex]
Work done by concrete, W = Force, F × Distance, d
- [tex]\displaystyle Force, \, F = \mathbf{\frac{Work, \, W}{Distance, \, d}}[/tex]
By the law of conservation of energy, we have;
The work done by the concrete, W = The kinetic energy, K.E. given by the safe
W = K.E. = 1,760,000 J
The effect of the work = The change in the height of the safe
Therefore;
The distance, d, over which the force of the concrete is exerted = The change in the height of the safe = 0.06 m
d = 0.06 m
Therefore;
[tex]\displaystyle The \ force \ of \ the \ concrete, \, F = \frac{1,760,000\, J}{0.06 \, m} = 29,333,333. \overline 3 \, N = 29.\overline 3 \ MN[/tex]
- The force of the concrete on the safe = [tex]\underline{29.\overline 3 \ MN}[/tex]
Second part:
The gravitational force of the Earth on the safe, W = The weight of the safe
W = Mass, m × Acceleration due to gravity, g
W = 2,200 kg × 9.81 m/s² ≈ 21,582 N
The ratio of the force exerted by the concrete to the weight of the safe is found as follows;
[tex]\displaystyle Ratio \ of \ forces = \frac{29.\overline 3 \times 10^6 \, N}{21,582 \, N} = \frac{4,000,000}{2,943} \approx \mathbf{1359.16}[/tex]
- The force exerted by the concrete is approximately 1,359.16 times the weight of the safe.
Learn more here:
https://brainly.com/question/21060171
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
