Answer:
Explanation:
The correct answer is Metabolic alkalosis (D). A pH of 7.48 shows slight alkalinity, this normal concentration of Co2 in the blood ranges from 35 mmHg (millimetre Mercury) to 45 mmHg and the normal HCo3 ( Hydrogen trioxo carbonate ion) concentration ranges from 22mEq/L to 26mEq/L.
Therefor the patients pH level is high the Co2 level is normal and the HCo3 level is high. Hence, Metabolic alkalosis
Answer: (a) t1 = omega1/alpha
(b) theta1 = 1/2 * alpha*theta1^2
(c) t2 = omega2/5*alpha
Explanation: see attachment
Answer:
Milliliters to Ounces Conversions
some results rounded
mL - fl oz
200.00 6.7628
200.01 6.7631
200.02 6.7635
200.03 6.7638
200.04 6.7642
200.05 6.7645
200.06 6.7648
200.07 6.7652
200.08 6.7655
200.09 6.7658
200.10 6.7662
200.11 6.7665
200.12 6.7669
200.13 6.7672
200.14 6.7675
200.15 6.7679
200.16 6.7682
200.17 6.7686
200.18 6.7689
200.19 6.7692
200.20 6.7696
200.21 6.7699
200.22 6.7702
200.23 6.7706
200.24 6.7709
mL fl oz
200.25 6.7713
200.26 6.7716
200.27 6.7719
200.28 6.7723
200.29 6.7726
200.30 6.7729
200.31 6.7733
200.32 6.7736
200.33 6.7740
200.34 6.7743
200.35 6.7746
200.36 6.7750
200.37 6.7753
200.38 6.7757
200.39 6.7760
200.40 6.7763
200.41 6.7767
200.42 6.7770
200.43 6.7773
200.44 6.7777
200.45 6.7780
200.46 6.7784
200.47 6.7787
200.48 6.7790
200.49 6.7794
mL fl oz
200.50 6.7797
200.51 6.7800
200.52 6.7804
200.53 6.7807
200.54 6.7811
200.55 6.7814
200.56 6.7817
200.57 6.7821
200.58 6.7824
200.59 6.7828
200.60 6.7831
200.61 6.7834
200.62 6.7838
200.63 6.7841
200.64 6.7844
200.65 6.7848
200.66 6.7851
200.67 6.7855
200.68 6.7858
200.69 6.7861
200.70 6.7865
200.71 6.7868
200.72 6.7872
200.73 6.7875
200.74 6.7878
mL fl oz
200.75 6.7882
200.76 6.7885
200.77 6.7888
200.78 6.7892
200.79 6.7895
200.80 6.7899
200.81 6.7902
200.82 6.7905
200.83 6.7909
200.84 6.7912
200.85 6.7915
200.86 6.7919
200.87 6.7922
200.88 6.7926
200.89 6.7929
200.90 6.7932
200.91 6.7936
200.92 6.7939
200.93 6.7943
200.94 6.7946
200.95 6.7949
200.96 6.7953
200.97 6.7956
200.98 6.7959
200.99 6.7963
Explanation:
Explanation:
The given data is as follows.
k = 130 N/m,
= 17 cm = 0.17 m (as 1 m = 100 cm)
mass (m) = 2.8 kg
When the spring is compressed then energy stored in it is as follows.
Energy = 
Now, spring energy gets converted into kinetic energy when the box is launched.
So,
= 
= 

= 1.34
v = 1.15 m/sec
Now,
Frictional force = 
= 
= 4.116 N
Also, Kinetic energy = work done by friction
1.8515 =
d = 0.449 m
Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.
Answer: To answer this question, we will need the following equation: SPEED = DISTANCE/TIME (A multiplication and division triangle will be shown)i) The speed of the car is calculated by doing 100 metres/ 20 seconds which gives us 5 metres per second. ii) Rearranging the equation earlier, we can make the distance the subject of the equation so that we get SPEED x TIME = DISTANCE. We worked out the speed and the time was given as 1 minute 40 seconds but we cannot plug in the numbers yet as the time has to be converted to units of seconds (because our speed is in meters per second). 1 minute 40 seconds = 60 seconds + 40 seconds = 100 secondsWe then plug in the numbers to get the distance travelled = 5 metres per second x 100 seconds = 500 metres.
Explanation: