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Explanation:
Let the east direction to be i^ and north direction be j^.
Thus displacement of the man S=8i^+6j^
So, magnitude of displacement ∣S∣=62+82=100=10 m
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
Answer:
Answered
Explanation:
The radius of curvature of the mirror R = 20 cm
then the focal length f = R/2 = 10 cm
(a) From mirror formula
1/f = 1/di + /1do
then the image distance
di = fd_o / d_o - f
= (10)(40) / 40-10
= 30.76 cm
since the image distance is positive so the image is real
ii) when the object distance d_0=20 cm
di = 10×20/ 20-10
= 20
Hence, the image must be real
iii)when the object distance d_0 = 10
di = 10×10 / 10-10 = ∞ (infinite)
the image will be formed at ∞
here also image will be real but diminished.
B. 1/2p
From the relation
P1V1=P2V2
P2=P1V1/V2
If everything equals one
P2=1*1/2(1)
That would equal 1/2
I hope this helps
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