<u>Explanation:</u>
s = ? u = 0m/s v = ? a = 5m/s² t = 2s
v = u + at
= 0 + (5 x 2)
= 10 m/s
s = ut + 1/2 at²
= (0)(2) + x
x 5 x 2²
= 10 m
Hope this helps!
Answer:
a) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
Answer:
0.076 m/s
Explanation:
Momentum is conserved:
m v = (m + M) V
(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V
V = 0.076 m/s
After catching the puck, the goalie slides at 0.076 m/s.
Answer:
Part a)

Part B)
percentage increase is
%
Explanation:
Part a)
As we know that the beat frequency is

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase
So we have


Part B)
percentage increase in the tension of the string will be given as


now we have

so we have


so we have

percentage increase is

I believe it would be a musical note