By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span>
=>v¹ = 2.25 m/s (East) </span>
<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let be the pressure at a point.
Let be the density fluid at a point.
Let be the velocity of fluid at a point.
Bernoulli's equation states that for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let be the pressure of a point just above the wing.
Let be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.
So,
Force is given by the product of pressure difference and area.
Given that area is .
So,lifting force is
Answer:
(A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Explanation:
Given that,
Fringe width d = 0.5 mm
Wavelength = 589 nm
Distance of screen and slit D = 1.5 m
Distance of bright fringe y = 1 cm
(A) We need to calculate the order of the bright fringe
Using formula of wavelength
Put the value into the formula
(B). We need to calculate the width of the bright fringe
Using formula of width of fringe
Put the value in to the formula
Hence, (A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Answer:
2.87 km/s
Explanation:
radius of planet, R = 1.74 x 10^6 m
Mass of planet, M = 7.35 x 10^22 kg
height, h = 2.55 x 10^6 m
G = 6.67 x 106-11 Nm^2/kg^2
Use teh formula for acceleration due to gravity
g = 1.62 m/s^2
initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0
Use third equation of motion
0 = v² - 2 x 1.62 x 2.55 x 10^6
v² = 8262000
v = 2874.37 m/s
v = 2.87 km/s
Thus, the initial speed should be 2.87 km/s.