Answer:
In terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
Explanation:
Stone 1:
vi = 10 m/s
vix = vi*Cos ∅ = (10 m/s)*Cos 30° = 8.66 m/s = vx
viy = vi*Sin ∅ = (10 m/s)*Sin 30° = 5 m/s
vy = viy - g*t = (5 m/s) - (9.8m/s²)*(1 s) = -4.8
then
v = √(vx²+vy²) = √((8.66)²+(-4.8)²) = 9.90 m/s
Δv = v - vi = 9.902 m/s - 10 m/s
⇒ Δv = -0.098 m/s
Stone 2:
vi = 10 m/s
v = vi + g*t = (10 m/s) + (9.8m/s²)*(1 s) = 19.8 m/s
Δv = v - vi = (19.8 m/s) - (10 m/s)
⇒ Δv = 9.8 m/s
Stone 3:
vi = 0 m/s
v = g*t = (9.8m/s²)*(1 s) = 9.8 m/s
Δv = v - vi = (9.8 m/s) - (0 m/s)
⇒ Δv = 9.8 m/s
Finally, in terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
Answer:
Explanation:
given,
velocity factor = 4
Cross-sectional area of venules(A₁) = 10 cm²
cross sectional area of capillaries(A_2) = ?
continuity equation = Q = AV
now,
hence, the area of capillaries is equal to 2.5 cm₂
Answer:
A: 15 m/s
B: 900 Meters
Explanation:
A. Average speed is equal to final velocity plus initial velocity divided by two
Va= (20+10)/2 = 15 m/s
B.
distance = speed × time
D= 15 m/s * 60s = 900 Meters
Answer:
Stephen hawking was born in January 8, 1942 in Oxford Eand. He received his undergraduate degree from the university of Oxford and later earned a PhD from the university of Cambridge.
Explanation: Stephen hawking was a theoretical physicist who was known for his work on black holes and nature of the universe.
Answer:
<h3>The answer is 0.92 m/s²</h3>
Explanation:
To find the acceleration of an object given it's initial and final velocity and time taken we use the formula
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question
v = 39 m/s
u = 27 m/s
t = 13 s
We have
We have the final answer as
<h3>0.92 m/s²</h3>
Hope this helps you