Answer:
I'm pretty sure its B and C
Explanation:
B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon
C bc mass is the same - you can't change it
Answer:
a)-1.014x
J
b)3.296 x
J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x
x 47 x 110/ 3.4 => -1.014x
J
b) at d= 0.8m (3.4-2.6) and
=-1.014x
J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+
=
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore
is zero
=
+ 
The final potential energy is
= - GMaMb/d
Solving for '
'
=
+ GMaMb/d => -1.014x
+ 6.67 x
x 47 x 110/ 0.8
= 3.296 x
J
Answer:
okay
Explanation:
(wish i could help but im just answering for points)
Transform boundary
is the answer from stemscopes
Answer:
a) f ’’ = f₀
, b) Δf = 2 f₀ 
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo 
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ 