Young athlete has a mass of 42 kg one day there is no wind shear and hundred metre race in 14.2 second a sketch graph not in ske
tch showing her speed during the race is
icalculate the acceleration of the athlete during the first 3 seconds of the race
ii the escalating for of the athlete during the first 3 seconds of the race
iii the with whicj she crosses the finish line.
icalculate the acceleration of the athlete during the first 3 seconds of the race
ii the escalating for of the athlete during the first 3 seconds of the race
iii the with whicj she crosses the finish line.
2 answers:
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Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity,
the angle of the slope
is the frictional force, with
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
Answer:
<h2>13N</h2>Explanation:
<em>Kindly see attached file for your reference</em>
Step one:
given data
the horizontal component of the force= 12N
the vertical component of the force= 5N
The dashed arrow represents the hypotenuse of the triangle, hence the resultant of the force system.
By implication of this, we will use the Pythagoras theorem to solve for the resultant force
Step two:
Answer:
Explanation:
Mass of the cable car, m = 5800 kg
It goes 260 m up a hill, along a slope of
Therefore vertical elevation of the car =
Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.
Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.
Hence, total change in energy = mgh =
where, g = acceleration due to gravity
h = height/vertical elevation