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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>
mass = 4 kg
velocity = 5 m/s^2
<h3 /><h3>Orange truck:</h3>Mass= 2kg
Velocity = 7m/s^2
<h3 /><h3>Grey Car:</h3>mass = 6 kg
velocity = 4m/s^2
<h3 /><h3>Green Car:</h3>Mass = 8 kg
Velocity = 3 m/s^2
<h2>_____________________________________</h2><h2>SOLUTION:</h2>By the equation of kinetic energy,
K.E =
Where,
K.E is kinetic energy
m is mass
v is velocity
<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>
Directly substitute the variables in the equation,
K.E =
Simplify the equation,
K.E = 50 J
<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>
Directly substitude the variable in the equation,
K.E =
Simplify the equation,
K.E = 48J
<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />Substitute the variables in the equation,
K.E =
Simplify the Equation,
K.E = 36J
<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />Substitute the variable,
K.E =
Simplify the equation,
K.E = 49J
<h2>_____________________________________</h2>As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />
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