How much work can be extracted from 5.00 × 10^4 J of heat when 2.00 × 10^4 J of heat is exhausted? A. - 7.00 × 10^4 J B. - 3.00
1 answer:
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Answer:
c = 4,444.44
Explanation:
You have the following expression for the acceleration of the projectile:
(1)
s: distance to the ground of the projectile
To find the value of the constant c you use the following formula:
(2)
vo: initial velocity = 0 m/s
v: final speed = 200 m/s
Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m
You replace the expression (1) into the expression (2):
You do the constant c in the last equation, then you replace the values of v, s and Δs:
<h2>Answer:</h2>
Torque = <em>2.05 x 10²⁸ Nm</em>
Energy = <em>3.54 x 10³³ J</em>
Average power = <em>1.02 x 10²⁸ W</em>
<h2>Explanation:</h2>(a) Torque (τ) is the rotational effect of a given force.
It is given by
τ = I x α -------------(i)
Where;
I = rotational inertia of the object
α = angular acceleration of the object.
In this case, the object is the Earth. Therefore,
I = 9.71 x 10³⁷ kg m²
α = ω / t
Where;
ω = angular velocity of earth = 2π rad / day
<em>Since </em>
<em>1 day = 24 hours and 1 hour = 3600seconds</em>
<em>1 day = 24 x 3600 seconds = 86400seconds</em>
<em>=> ω = 2π rad / 86400seconds</em>
<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>
<em />
t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds
=> α = ω / t
=> α = 7.29 × 10⁻⁵ / 345600
=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)
=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)
=> α = (7.29 × 10⁻¹⁰) / (3.456)
=> α = 2.11 × 10⁻¹⁰ rad/s²
Now substitute the values of I and α into equation (i)
τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰
τ = 9.71 x 10²⁷ x 2.11
τ = 20.5 x 10²⁷ Nm
τ = 2.05 x 10²⁸ Nm
(ii) The energy (rotational energy) E is given by;
E = x I x ω
E = x 9.71 x 10³⁷ x 7.29 × 10⁻⁵
E = 35.4 x 10³² J
E = 3.54 x 10³³ J
(iii) The average power P, is given by;
P = E / t
P = 3.54 x 10³³ / 345600
P = 1.02 x 10²⁸ W