Do you have a picture of the diagram that I could view?
Answer:
(A) v = 14.8m/s
Explanation:
(A) V = sqrt(k/m) × A = sqrt(22/0.1) × 0.29 =14.8m/s.
The
heating coils in you toaster during the first five seconds after you turn the
toaster on is Δ K = W + Q.
<span>Your automobile just before you fill
it with gas until you pull away from the gas station at speed v is
ΔK
+ ΔU + ΔEint = W + Q + TMW + TMT</span>
<span>your
body while you sit quietly and eat a peanut butter and jelly sandwich for lunch
is
ΔEint = Q + TET + TER</span>
<span>
your home during five
minutes of a sunny afternoon while the temperature in the home remains fixed is
ΔU = W + Q + TMW + TMT</span>
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is, 
Tangential speed of the skater is, 
Mass of the skater is, 
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, 
. This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
