Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.
From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.
Even though the favor that was asked by Dr. Hsu was prohibited by AAMA, it should be noted that the final part of the favor about faxing the report to the internist would fall within AAMA guidelines.
In conclusion, the best way that Kayla can respond to Dr. Hsu is to decline doing the favor.
Read related link on:
Answer:
2.19 N/m
Explanation:
A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:
T = 2π√(m/K)
Where T is the period, m is the mass (in kg), and K is the damping constant. So:
2.4 = 2π√(0.320/K)
√(0.320/K) = 2.4/2π
√(0.320/K) = 0.38197
(√(0.320/K))² = (0.38197)²
0.320/K = 0.1459
K = 2.19 N/m
Answer:
Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous planet increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers
Explanation:
A. 50 kilometers
Btu/(lb-°F) J/(g-°C i mean this is the correct answer
