Answer:
The constant here is the study outline
Explanation:
In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.
Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.
NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.
Answer:
Answer D
Explanation:
They need to be checked monthly
Answer:
T_warm = 47.22 C
Explanation:
Using energy balance for the system:
m_1*h_1 + m_2*h_2 = m_3*h3 ... Eq1
h_i = c_p. T_i ... Eq 2
m_1 + m_2 = m_3 ... steady flow system (Eq 3)
Substitute Eq 2 and Eq3 in Eq1
m_3 = 0.8 + 1 = 1.8 kg/s
(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3
T_3 = 355.3 / (1.8*4.18) = 47.22 C
This question is incomplete, the complete question is;
A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle.
Determine the velocity of the flowing water jet if the blade is assumed to be frictionless.
Answer: the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
Explanation:
Given that;
Force Ft = 1040 N
diameter d = 2.5 cm = 0.025 m
we know that; force acting on Hemispherical plate is;
Ft = 2δav²
where
a is area = π/4(0.025)²
δ is density of water = 1000 kg/m³
v is velocity = ?
now we substitute
1040 = 2 × 1000 × (π/4(0.025)²) × v²
1040 = 0.9817v²
v² = 1040 / 0.9817
v² = 1059.3867
v = √1059.3867
v = 32.5482 ≈ 32.55 m/s
Therefore the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
