Answer:
V = 365643.04 m/s
Explanation:
mass of the sun = 1.99 x 10^{30} kg
mass of M1 = mass of M2 = 6.95 solar mass = 6.95 x 1.99 x 10^{30} = 13.8305x 10^{30} kg
orbital period of each star (T) = 2.20 days = 2.20 x 24 x 60 x 60 =190,080 s
gravitational constant (G) = 6.67 x 10^{-11} N m2/kg2
orbital speed (V) = 
we need to find the orbital radius (r) before we can apply the formula above and we can get it from Kepler's third law,
x k
where
- k =
(take note that π is shown as
)
making r the subject of the formula we now have
(take note that π is shown as
)

r = 1.38 x 10^{10} m
Now that we have the orbital radius (r) we can substitute all required values into the formula for orbital speed
orbital speed (V) = 

V = 365643.04 m/s
I think the answer you or looking for is miter saw please tell me if I’m wrong
Answer:
80×5×10=4000J
so therefore, work done on the body is 4000J
ANSWER:
(a) 1036 N
(b) -1036 N
(c) 2590 N
STEP-BY-STEP EXPLANATION:
Given:
Mc = 1400 kg
Mt = 560 kg
a = 1.85 m/s^2
(a)
Force by car on trailer:

(b)

(c)
Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm
Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as
y = R×(mλ/d)
Where y = distance between nth fringes and the center fringe.
m = order of fringe
λ = wavelength of light = 589nm = 589×10^-9m
R = distance between slits and screen = 1.0m
d = distance between slits = 0.25mm = 0.00025m
For distance between the first dark fringe and the center fringe.
This implies that m = 1
y = 1 × 589×10^-9 × 1/0.00025
y = 589×10^-9/0.00025
y = 2,356,000 × 10^-9
y = 2.356 × 10^-3 m = 2.356mm
For the second dark fringe, this implies that m = 2
y = 1 × 2 × 589×10^-9/0.00025
y = 1178 × 10^-9 /0.00025
y = 4,712,000 × 10^-9
y = 4.712 × 10^-3 m = 4.712mm