Answer:
<h3>idl low presure but de 2.518 has a 6 divide 8 equal di. ko alam?</h3>
Answer:
v = 0.059 m/s
Explanation:
To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:
(1)
m: mass of the ball = 0.400kg
M: mass of Olaf = 75.0 kg
v1i: initial velocity of the ball = 11.3m/s
v2i: initial velocity of Olaf = 0m/s
v: final velocity of Olaf and the ball
You solve the equation (1) for v and replace the values of all variables:
Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s
Answer: 1175 J
Explanation:
Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."
Given
Spring constant, k = 102 N/m
Extension of the hose, x = 4.8 m
from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m
Work done =
W = 1/2 k [x(i)² - x(f)²]
Since x(f) = 0, then
W = 1/2 k x(i)²
W = 1/2 * 102 * 4.8²
W = 1/2 * 102 * 23.04
W = 1/2 * 2350.08
W = 1175.04
W = 1175 J
Therefore, the hose does a work of exactly 1175 J on the balloon
1. A. medium
2. B. matter
3. B. straight lines
im sorry i couldnt help for number 4 but hope this helps
Answer: M^-1 L^-3T^4A^2
Explanation:
From coloumb's law
K = q1q2 / (F × r^2)
Where;
q1, q2 = charges
k = constant (permittivity of free space)
r = distance
Charge (q) = current(A) × time(T) = TA
THEREFORE,
q1q2 = (TA) × (TA) = (TA)^2
Velocity = Distance(L) / time(T) = L/T
Acceleration = change in Velocity(L/T) / time (T)
Therefore, acceleration = LT^-2
Force(F) = Mass(M) × acceleration (LT^-2)
Force(F) = MLT^-2
Distance(r^2) = L^2
From ; K = q1q2 / (F × r^2)
K = (TA)^2 / (MLT^-2) (L^2)
K = T^2A^2M^-1L^-1T^2 L^-2
COLLEXTING LIKE TERMS
T^2+2 A^2 M^-1 L^-1-2
M^-1 L^-3T^4A^2
