v = v₀ + at
v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time
Given values:
v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s
Plug in and solve for v:
v = 0 + 9.81(3)
v = 29.4m/s
V = IR
By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.
So i believe the answer would be : The string will remain lit
hope this helps
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F
× t = m( v - v₀ )
we substitute
F
× 0.00265 = 1.75( 0 - 5.25 )
F
× 0.00265 = 1.75( - 5.25 )
F
× 0.00265 = -9.1875
F
= -9.1875 / 0.00265
F
= -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F
= - F
we substitute
F
= - ( -3466.98 N )
F
= 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct
Answer:
Magnetic field, B = 0.199 T
Explanation:
It is given that,
Radius of circular loop, r = 11.7 cm = 0.117 m
Magnetic flux through the loop, 
The magnetic flux linked through the loop is :


Here, 

or


B = 0.199 T
So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.