Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as
Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²
~Formula: Voltage= current• resistance
(V= Ir)
~Using this formula, plug in the numbers from the equation into the formula
~5=25i
~Now you have a one-step equation
~Divide by 25 on both sides and you should get your answer:
~I= 0.2 (which means current is 0.2)
Answer:
total momentum = 8.42 kgm/s
velocity of the first cart is 3.660 m/s
Explanation:
Given data
mass m1 = 2.3 kg
mass m2 = 1.5 kg
final velocity V2 = 4.9 m/s
final velocity V3 = - 1.9 m/s
to find out
total momentum and velocity of the first cart
solution
we know mass and final velocty
and initial velocity of second cart V1 = 0
so now we can calculate total momentum that is m1 v2 + m2 v2
total momentum = 2.3 ×4.9 + 1.5 ×(-1.9)
total momentum = 8.42 kgm/s
and
conservation of momentum is
m1 V + m2 v1 = m1 v2 + m2 v3
put all value and find V
2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)
V = 8.42 / 2.3
V = 3.660 m/s
so velocity of the first cart is 3.660 m/s
Answer:
a) t = 0.528 s
b) D = 1.62 m
Explanation:
given,
speed of the baseball = 3.75 m/s
angle made with the horizontal = 35°
height of the roof edge = 2.5 m
using equation of motion
4.9 t² + 2.15 t - 2.5 = 0
on solving the above equation
t = 0.528 s
b) D = v cos θ × t
D = 3.75 × cos 35° ×0.528
D = 1.62 m
Answer:
It corresponds to 1mm-10 mm range.
Explanation:
- Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
- As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:
- Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:
