Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Answer:
-22.2 m/s²
Explanation:
The equation for position x for a constant acceleration a, time t and initial velocity v₀, initial position x₀:
(1) 
For rocket A the initial and final position: x = x₀= 0. Using these values in equation 1 gives:
(2) 
Solving for time t:

(3) 
The times for both rockets must be equal, since they start and end at the same location. Using equation 3 for rocket A and B gives:
(4) 
Solving equation 4 for acceleration of rocket B:
(5) 
Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83 Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640 m^3/s for in and 1000 m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
B.) <span>The amp is the unit for "Current"
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