Answer:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
Explanation:
Okay soo-
Given-
u = 60 km/hr = 60×1000/3600=50/3 m/s
t = 20 s
s = 250 m
a = ?
v = ?
Solution -
Here, acceleration is uniform.
(a) According to 2nd kinematics equation,
s = ut + ½at^2
250 = 50/3 ×20 + 0.5×a×20×20
250-1000/3=200a
(750-1000)/3=200a
a = -250/(3×200)
a = -5/12
a = 0.4167 m/s^2
The required uniform acceleration of the car is 0.4167 m/s^2.
(b) According to 1st kinematics equation
v = u + at
v = 50/3 + (-5/12)×20
v = 50/3-25/3
v = 25/3
v = 8.33 m/s
The speed of the car as it passes the traffic light is 8.33 m/s.
Good luck!
Answer:
Area of Circle = 78.5398
Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft
Volume of Sphere = 33.5103 ft
Explanation:
Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.
r=5; % define r as 5
a=pi*r^2;% calculate the area of the circle
AreaOfCircle=a
r=10; % define r and 10 ft
sa=4*pi*r^2; %Calculate the surface area of the sphere
SphereSurfaceArea=sa
r=2;% define r as 2 ft
vs=(4/3)*pi*r^3;% Calculate the volume of the sphere
VolumeShere=vs