<span>D. Subject matter
</span>Hat is most remarkable about the paintings of Pieter Breugel the Elder?NOT:
A. Innovative brush strokes
B. Medium used
<span>C. Daring compositional features</span>
Write each force in component form:
<em>v </em>₁ : 50 N due east → (50 N) <em>i</em>
<em>v</em> ₂ : 80 N at N 45° E → (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )
The resultant force is the sum of these two vectors:
<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>
Its magnitude is
|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N
and has direction <em>θ</em> such that
tan(<em>θ</em>) = (56.5 N) / (106.5 N) → <em>θ</em> ≈ 28.0°
i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)
For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0.
</span>
y(x,t) = A sin(kx-omega t)
d{y(x,t)}/d{x} = A k cos(kx - omega t)
Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is 
The value for static friction is 
Explanation:
From the question we are told that
The mass of the clock is 
The first horizontal force is 
The second horizontal force is 
Generally the static frictional force is equal to the first horizontal force
So
=> 
=>
Generally the kinetic frictional force is equal to the second horizontal force
So
Answer:
Explanation:
the same amount of time in both halves of the circle
