Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
John carry the heaviest load.
<h3>How to find out who is carrying the heavy load?</h3>
Write down given data from questions:
Board=510cm X 510mm.
Cylinder head with dimensions=43cm X 250mm.
Cylinder lies across the board 210cm from john.
Find out: Who is carry the heaviest load?
Calculation:
We assume that mass of cylinder head = x kg
Then weight=x x 9*81
W=9.81x Newton.
Weight per unit length= Weight/Total leanth
Weight per unit length= 9.81x/43
(w/l)=0.23x N/cm
From equation contition: 
(210+21.5)




Therefore 
To learn more about mass per unit length, refer to:
brainly.com/question/24180692
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Answer:
mass consumed by 235U each day = 2 kg
Explanation:
electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV
= 6.24151× 10²¹ MeV/s
thermal energy = 0.420 * 250 = 105 MeV

= 5.94 × 10¹⁹ fission/second
=5.94 × 10¹⁹× 24 × 60 ×60)
= 5.13 × 10²⁴ fission/day
mu = 235.04393 × 1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg
M = mu ×5.13 × 10²⁴
= 390.1729× 10⁻²⁷ ×5.13 × 10²⁴
M = 2 kg(approx.)
mass consumed by 235U each day = 2 kg
Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC


s = 68.94 m

3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.