Answer:
a. 450 b. 10078.9097
Step-by-step explanation:
a. go with the formula W=FD
F= force
W= work done
D= distance
b. go with the formula F=mgcos(x)
M= mass
G= gravity constant (9.8)
then get to the normal formula of W=FD after getting the Force
Answer:
c
Step-by-step explanation:
Answer:It should be x=1 not sure thought.
Step-by-step explanation:
Answer:
0.6 °C/min
Step-by-step explanation:
The relationship between rates and movement is ...
time = distance/speed
Here, the "distance" is measured in °C, and the "speed" is the rate of change of temperature.
For the first half of the heating, the time required is ...
(50°C -0°C)/(1.5 °C/min) = 50/(3/2) min = 100/3 min
For the second half of the heating, the time required is ...
(100°C -50°C)/(4/10 °C/min) = 50/(4/10) = 125 min
Then the total time is ...
((100/3) +125) min = (475/3) min
And the average rate of temperature increase is ...
total temperature change / total time
= (100°C -0°C)/(475/3 min) = 300/475 °C/min = 12/19 °C/min ≈ 0.6 °C/min
Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
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You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
