Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
Answer is B.
In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.
Hope it helped you.
-Charlie
Answer:
The frictional force needed to overcome the cart is 4.83N
Explanation:
The frictional force can be obtained using the following formula:
where 
is the coefficient of friction = 0.02
R = Normal reaction of the load = 
= 
= 
Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.
F = 4.83 N
Hence, the frictional force needed to overcome the cart is 4.83N
Answer:
0.0675 seconds
Explanation:
From the question,
We apply newton's second law of motion
F = m(v-u)/t.................... Equation 1
Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.
make t the subject of the equation
t = m(v-u)/F................... Equation 2
Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)
Substitute these value into equation 2
t = 180(0-6.0)/-1600
t = -1080/-1600
t = 0.0675 seconds.
