There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"
Answer:
100 cc
Explanation:
Heat released in cooling human body by t degree
= mass of the body x specific heat of the body x t
Substituting the data given
Heat released by the body
= 70 x 3480 x 1
= 243600 J
Mass of water to be evaporated
= 243600 / latent heat of vaporization of water
= 243600 / 2420000
= .1 kg
= 100 g
volume of water
= mass / density
= 100 / 1
100 cc
1 / 10 litres.
Answer:
a. a=33.34ms⁻², V=164.4m/s
Explanation:
Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.
Below are the data given
Distance, s=404.5m,
time taken,t=4.922secs
Using the equation
S=ut+1/2at²
where u is the initial velocity and u=0
Making the acceleration the subject of the formula, we arrive at
a=2s/t²
a=(2*404.5)/4.922²
a=33.34ms⁻².
To determine the velocity, we use
V=u+at
V=0+33.34ms⁻² *4.922sec
V=164.4m/s
Answer:
7.82 s
Explanation:
Given:
Δy = 300 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.82 s
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
