Answer: c. A person who gains unauthorized access to digital data
Explanation:
Answer:
Explanation:
Let the time required for acceleration a₁ and deceleration a₂ be t₁ and t₂ .
Since final velocity during acceleration and initial velocity during deceleration are same
a₁ t₁ = a₂ t₂
5t₁ = 2 t₂ ------------------------------------------ ( 1 )
Distance travelled during acceleration = 1/2 a₁t₁²
= 1/2 x 5 x t₁² = 2.5 t₁²
Distance travelled during deceleration = 1/2 a₂t₂²
= 1/2 x 2 x t₂² = t₂²
Total distance travelled = 2 miles = 2 x 1760 x 3 ft = 10560
2.5 t₁² + t₂² = 10560
2.5 ( 2t₂ / 5 )² + t₂² = 10560
.4 t₂² + t₂² = 10560
1.4 t₂² = 10560
t₂ = 86.85 s
t₁ = 2t₂ / 5 = 34.75 s
t₁ + t₂ = 121.6 = 122 s
Total time taken = 122 s .
maximum velocity = a₁t₁
= 5 x 34.75 = 173.75 = 174 m/s .
The equatorial radius of the earth is
r = 6378 km = 6378 x 10³ m
The earth makes 1 revolution in 24 hours.
The angular velocity is
ω = (2π rad)/(24*3600 s) = 7.2722 x 10⁻⁵ rad/s
The tangential velocity (linear velocity) at a point on the equator is
v = rω
= (6378 x 10³ m)*(7.2722 x 10⁻⁵ rad/s)
= 463.8 m/s
Answer: 463.8 m/s
Answer:
5.865 μs
Explanation:
t₀ = Time taken to decay a muon = 2.20 μs
c = Speed of Light in vacuum = 3×10⁸ m/s
v = Velocity of muon = 0.927 c
t = Lifetime observed
Time dilation
∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs
