At time=0s, the object is at the 21.0-meter position along the roadway. Where is the object at time = 10 s?
1 answer:
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Answer:
speed = 90.7934 km/hr
bearing = 88.55°
Explanation:
given data
plane speed v = 112 km/hr
bearing = 81°
θ = 90 - 81 = 9°
wind speed w = 28 km/hr
bearing (θ2) = 225°
to find out
ground speed and the bearing of the plane
solution
we get here resultant ground speed of plane that is
speed = [ v cos(θ) + w cos(θ2)] i + [ v sin(θ) + w sin(θ2)] j
speed = [112 cos(9) + 28 cos(225)] i + [112 sin(9) + 28 sin(225)] j
speed = 90.822 i - 2.278 j
| speed | =
| speed | = 90.7934 km/hr
and
tan(θ) =
θ = 1.4367°
bearing of the plan will be
bearing = 90 - 1.4367
bearing = 88.55°
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ + t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Answer:
The answer to the question is;
The net force acting on the dam is 73,001,115 N.
Explanation:
To solve he question, we note that
Pressure = Force/Area and the formula for pressure is
Pressure = hρg
While force, F= Pressure × Area
For a small area at the bottom of the dam we have
= hρg×dh×width
Where
w = width of the dam = 125 m
ρ = Density of water = 1.0×10³ m³
g = Acceleration due to gravity = 9.81 m/s²
h = Height of the dam = 11.0 m
Therefore force over the dam wall = =
= [ 11²/2×1×10³×9.81×125] - 0
= 73,001,115 N
The net force acting on the dam = 73,001,115 N.