consider the forces on mass m₁ on the incline plane :
parallel to incline , force equation is given as
T - m₁ g Sin30 = m₁ a
T = m₁ g Sin30 + m₁ a eq-1
consider the force on mass m₂ on the incline plane :
m₂ g - T = m₂ a
T = m₂ g - m₂ a eq-2
Using eq-1 and eq-2
m₂ g - m₂ a = m₁ g Sin30 + m₁ a
inserting the values
(2.3 x 9.8) - 2.3 a = (3.7 x 9.8) Sin30 + 3.7 a
a = 0.74 m/s²
What is the impulse on the skateboarder?
What is the average force on the skateboarder in each of these stops?
1) Impulse:
2) Strong force: -131.1 N, weak force: -44.7 N
Explanation:
1)
The impulse exerted on an object is equal to the change in momentum of the object itself.
Mathematically:
where
m is the mass of the object
u is the initial velocity
v is the final velocity
For the skateboarder in this problem, we have:
m = 53.6 kg
u = 4.45 m/s
v = 0 (it comes to a stop)
Therefore, the impulse is
Where the negative sign indicates that the direction is opposite to the motion of the object.
2)
The impulse is also equal to the product between the force applied and the duration of the collision:
where
I is the impulse
F is the average force
is the time during which the force is applied
The strong force is applied in a time of
Therefore this force is
The weak force is applied in a time of
So this force is
Learn more about impulse:
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Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s
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