The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.

= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>
Generally, the equation for the is mathematically given as

Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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Answer:
Explanation:
we humans have our own ify classification for celestial objects, most people are saddened that pluto is not a planet anymore altho it hasn't changed at all.
scientist say that if an object is going to be considered a planet it must fill in these three checkboxes:
You must be spherical, you must orbit a star, and you must have already cleared your path or debris.
Pluto fills in the first two boxes but it does orbit in the keyperbelt and there are 5 other objects just like it. this is why pluto has been dubbed a dwarf planet.

Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
R is proportional to the length of the wire:
R ∝ length
R is also proportional to the inverse square of the diameter:
R ∝ 1/diameter²
The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.
Calculate the scale factor due to the changed length:
k₁ = 2800/2700 = 1.037
Scale factor due to changed diameter:
k₂ = 1/(0.1/0.26)² = 6.76
Multiply the original resistance by these factors to get the new resistance:
R = R₀k₁k₂
R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76
R = 9850(1.037)(6.76)
R = 69049.682Ω
Round to the nearest hundredth:
R = 69049.68Ω