a) 6.52 m/s^2
b) 23.47 km/hr^2
a) v = v0 + at --> 9 = 0 + a(1.38) --> a = 6.52
b) 6.52 m/s2 * 60 s2 / 1 min2 * 60 min2 / 1 hr2 = 23.47 km/hr2
Answer:
so length of the fence is 250 feet
Explanation:
Given data
fencing (L) = 500 feet
enclose the garden = 3 sides
to find out
length of the fence
solution
we consider barn length x and width y
so we say x +2y = 500 ........1
we want to maximize length i.e xy = f
so we say
df/dx = ƛ × dL/dx
y = ƛ × 1
and
df/dy = ƛ × df/dy
x = ƛ × 2
from equation 1 we can say
2 ƛ + 2 ƛ = 500
so ƛ = 500/4 = 125
so that x = 2× 125 , y = 125
so x = 250 , y = 125
so length of the fence is 250 feet
<em>A straight piece of wire with a current I flowing through it is placed in a magnetic field</em>
<em>A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the string</em>
<em>A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the stringthere is a way</em>
Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
The answer is chicken nuggets
mark me brainliest
