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Svetach [21]
3 years ago
6

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s
Physics
1 answer:
Burka [1]3 years ago
8 0

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

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A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.
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Heat lost or gained, H = mc(θ₂ - θ₁) 
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C  (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444)  =  θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂      =  77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C .  Option C.


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