Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
, T is tension in the wire , m is mass per unit length of wire .
40.48 L = 
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s
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The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
Answer:
6 hours 15 minutes
Explanation:
On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed). 5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes. On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind. So the net (groundspeed) is 800 mph. Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.
Outbound flight 7 hours 9 minutes
Return flight 6 hours 15 minutes
