We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2). <span>
We want a ratio of 3000000 to 1, or 3 million to 1.
</span>
<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>
Work formula:
W = F * d
F 1 = 40 N, d 1 = 6 m;
F 2 = 30 N; d 2 = 6 m.
W ( Cindy ) = 40 * 6 = 240 Nm
W ( Andy ) = 30 * 6 = 180 Nm
The difference of their amounts if work:
240 Nm - 180 Nm = 60 nm
hope it helps!
Answer:
The magnitude of buoyancy force is equal to that of ball's weight.
Explanation:
Ball 1 is floating on water. Weight of ball 1 is Fg=m1g is acting vertically downward
Force of buoyancy FB = ρVdisg is acting vertically upward.
Net force acting on the ball is zero, FB=Fg
Answer
The magnitude of buoyancy force is equal to that of ball's weight.
Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 ×
m
separation L = 0.5 mm = 0.5 ×
m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 ×
m
λ = 2042 nm
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
