Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
(1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
(2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
Ec: Calories
You solve for Ec and convert the result to Cal:
The amount of Calories needed by the climber was 6220.56 kcal
<span>if the airplane moves at a speed of 25 m/s the mass is 855kg</span>
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
<span>step 1: energy required to heat coffee
E = m Cp dT
E = energy to heat coffee
m = mass coffee = 225 mL x (0.997 g / mL) = 224g
Cp = heat capacity of coffee = 4.184 J / gK
dT = change in temp of coffee = 62.0 - 25.0 C = 37.0 C
E = (224 g) x (4.184 J / gK) x (37.0 C) = 3.46x10^4 J
step2: find energy of a single photon of the radiation
E = hc / λ
E = energy of the photon
h = planck's constant = 6.626x10^-34 J s
c = speed of light = 3.00x10^8 m/s
λ = wavelength = 11.2 cm = 11.2 cm x (1m / 100 cm) = 0.112 m
E = (6.626x10^-34 J s) x (3.00x10^8 m/s) / (0.112 m) = 1.77x10^-16 J
step3: Number of photons
3.46x10^4 J x ( 1 photon / 1.77x10^-16 J) = 1.95x10^20 photons</span>
Answer:
A fair test.
Explanation:
Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.
In a fair test two or more things are compared.
It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.
The factor is called a variable.