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liq [111]
1 year ago
14

derric climbs onto his roof to hang Christmas light. If the roof is 4.3 meters from the ground and derrick gains 2730 J of poten

tial energy while climbing, what is his mass​
Physics
1 answer:
Lelu [443]1 year ago
4 0

Mass of Derrick at a height 4.3 m having 2730 joule of energy is 65 Kg.

<h3>What is the expression of gravitational potential energy near the earth surface?</h3>
  • Mathematically, gravitational potential energy near the earth surface= m×g×h
  • m= mass of Derrick, g= acceleration due to gravity, h= height at which Derrick present
  • Then, mass (m) of Derrick = potential energy/ (g×h)

<h3>What is the mass of Derrick, if he gains 2730 joule of energy at 4.3m above the ground?</h3>

Mass of Derrick= 2730/(9.8×4.3)

= 65 kg

Thus, we can conclude that the mass of Derrick is 65 Kg.

Learn more about the gravitational potential energy here:

brainly.com/question/26588957

#SPJ1

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A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary
Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

  • <em>mass of the bullet, m = 23 g = 0.023 g</em>
  • <em>speed of the bullet, u = 230 m/s</em>
  • <em>mass of the wood, m = 2 kg</em>
  • <em>final speed of the bullet, v = 170 m/s</em>
  • <em>coefficient of friction, μ = 0.15</em>

The final velocity of the wood after the bullet hits is calculated as follows;

m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s

The acceleration of the wood is calculated as follows;

\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2

The distance traveled by the wood after the bullet emerges is calculated as follows;

v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0
3 years ago
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
RideAnS [48]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

4 0
3 years ago
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