Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
RADIATION BELTS....... I think but it should be radiation belt
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb = 
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.
= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) = 
= 
= 
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g = 
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Answer:
23.52 m/s
Explanation:
The following data were obtained from the question:
Time taken (t) to reach the maximum height = 2.4 s
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =..?
At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)
was thrown as follow:
v = u – gt (since the rock is going against gravity)
0 = u – (9.8 × 2.4)
0 = u – 23.52
Collect like terms
0 + 23.52 = u
u = 23.52 m/s
Therefore, the rock was thrown at a velocity of 23.52 m/s.
The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m
</span>b. the displacement of car B - <span>65.5 m
c. average velocir</span>y of A
d. the average velocity of car B has the same.
