Answer:
The work done against gravity is 78.4 J
Explanation:
The work is calculated by multiplying the force by the distance that the
object moves
W = F × d, where W is the work , F is the force and d is the distance
The SI unit of work is the joule (J)
We need to find the work done against gravity when lowering a
16 kg box 0.50 m
→ F = mg
→ m = 16 kg, and g = 9.8 m/s²
Substitute these value in the rule
→ F = 16 × 9.8 = 156.8 N
→ W = F × d
→ F = 156.8 N and d = 0.50
Substitute these values in the rule
→ W = 78.4 J
<em>The work done against gravity is 78.4 J</em>
From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545
Answer:
Both are aquatic animals and are hunters
Explanation:
Yes, it can be unicellular and multicellular
Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula






A negative result has no sense, therefore the rowing speed in still water was 3 km/h