Answer:
T_ww = 43,23°C
Explanation:
To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:
E_in=E_out+E_loss
The energy associated to a current of fluid can be defined as:
E=m*C_p*T_f
So, applying the energy balance to the system described:
m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss
Replacing the values given on the statement, we have:
1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30 kJ/s
Solving for the temperature Tww, we have:
(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW
T_WW=43,23 °C
Have a nice day! :D
Answer:

Explanation:
When we shoot the dart upwards the time taken by the dart to go straight up and again come back is given as

here we can say


put t = 4.6 s then we have

Now in order to find the maximum range we can say

so in order to have maximum range we can say




Gravity is stronger the closer you get. It is D.
The answer is deposition/A. Please mark brainliest.
Answer:
(a) v1 = 21.6 m/s
(b) t = 51.25 s
Explanation:
Use kinematics equation
v1 = v0 + at
Given
v0 = 0 = initial velocity
a = 0.8 m/s^2 = acceleration
(a) t = 27 seconds
v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s
(b) v1 = 41 m/s
v1 = v0 + at
solve for t
t = (v1-v0)/a = (41-0)/0.8 = 51.25 s