Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
The answer is going to be element #29 Copper makes blue
Red:#38
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Hopes This Helps
<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities
sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s
cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s
Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s.
To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So
0.50 s * 14.7447368 m/s = 7.372368399 m
1.00 s * 14.7447368 m/s = 14.7447368 m
1.50 s * 14.7447368 m/s = 22.1171052 m
2.00 s * 14.7447368 m/s = 29.48947359 m
Rounding the results to 1 decimal place gives
0.50 s = 7.4 m
1.00 s = 14.7 m
1.50 s = 22.1 m
2.00 s = 29.5 m</span>
Answer:
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
Initial speed of the projectile is v and final speed is 0.5 v.
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
So, the height of the projectile above the ground is 
. Hence, this is the required solution.
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.
