Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J
Q = mcΔT
<span>q = 55.8g x 0.450J/gC x 23.5C </span>
<span>q = 590. J ................ to three significant digits
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Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
V=IR, therefore when resistance is constant the voltage and current are directly proportional