The formula we can use in this case is:
v = v0 + a t
where v is final velocity, v0 is initial velocity, a is
acceleration and t is time
So finding for v0:
v0 = v – a t
v0 = 43.7 – (2.5) 2.7
v0 = 36.95 m/s
I believe your answer is TRUE!
Hope this helps!:)
Given: Mass m = 44 Kg; Velocity v = 10 m/s
Required: Kinetic energy K.E = ?
Formula: K.E = 1/2 mv²
K.E 1/2 (44 Kg)(10 m/s)²
K.E = 2,200 Kg.m²/s²
K.E = 2,200 J Answer is A
Answer:
D. 2^(3/2)
Explanation:
Given that
T² = A³
Let the mean distance between the sun and planet Y be x
Therefore,
T(Y)² = x³
T(Y) = x^(3/2)
Let the mean distance between the sun and planet X be x/2
Therefore,
T(Y)² = (x/2)³
T(Y) = (x/2)^(3/2)
The factor of increase from planet X to planet Y is:
T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)
T(Y) / T(X) = (2)^(3/2)
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.
