In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
To get the solution you must need to draw a force triangle. Attach the head of the 60N north force arrow with the tail of the 60N east force arrow. The subsequent is the arrow connecting he tail and head of the two arrows.
You get a right angled triangle, and the resultant is (60^2 + 60^2) ^0.5 = 84.85 N or 85 N northeast.
Answer:
Explanation:
As we know that the ball is projected upwards so that it will reach to maximum height of 16 m
so we have
here we know that
also we have
so we have
Now we need to find the height where its speed becomes half of initial value
so we have
now we have
If this case could ever happen, the speed would follow from this formula:
with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."
The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)
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
