If you want to conduct an electrical current, which situation would produce a solution capable of this? A) Dissolving water in o
2 answers:
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Answer:13.2,41.4,82.8 ft
Explanation:
Given Person is 6.60 ft from the left hand side mirror
Since the focal length of plane mirror is infinity therefore image and object are equidistant from mirror
Distance of first image on left mirror is
i.e. image is 13.2 ft away from object
Second image
Now the right mirror forms the image of object at distance of 14.1 ft right from right mirror so its image is formed in left mirror at a distance of 34.8 from left mirror
so image distance from person is
Third image now form image of second image is formed on right mirror at a distance of 55.5 right from right mirror
and its mirror image is formed on left mirror at a distance of 76.2 left
from left mirror
Answer:
1.V= 640.48 m/s :total velocity in t= 5s
2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away
3. v =25m/s
4. s= (-1.5t³+26t ) m
Explanation:
1. Parabolic movement in the x-y plane , t=5s
V₀=638.6 m/s=Vx :Constant velocity in x
Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y
V= 640.48 m/s : total velocity in t= 5s
2.
x=v₀x*t
13=v₀x*t
13=17.49*t
t=13/17.49=0.743s : time for 13.0 m away
th=v₀y/g=11.44/9.8= 1,17s :time for maximum height
at t=0.743 sthe ball is going up ,then g is negative
y=v₀y*t - 1/2 *g¨*t²
y=11.44*0.743 -1/2*9.8*0.743²
y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away
3. s = (1t3 + -5t2 + 3) m
v=3t²-10t=3*25-50=75-50=25m/s
at t=0, s=3 m
at t=5s s=5³-5*5²+3
4. a = (-9t) m/s2
a=dv/dt=-9t
dv=-9tdt
v=∫ -9tdt
v=-9t²/2 + C1 equation (1)
in t=0 , v₀=26m/s ,in the equation (1) C1= 26
v=-9t²/2 + 26=ds/dt
ds=( -9t²/2 + 26)dt
s= ∫( -9t²/2 + 26)dt
s= -9t³/6+26t+C2 Equation 2
t = 0, s = 0 , C2=0
s= (-9t³/6+26t ) m
s= (-1.5t³+26t ) m