Answer:
0.025 m
Explanation:
From the question,
Applying Hook's law
F = ke................... Equation 1
Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.
make e the subject of the equation
e = F/k....................... Equation 2
Given: F = 10 N, e = 395 N/m
Substitute these values into equation 2
e = 10/395
e = 0.025 m
Answer:
Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
v2f=v2i−2⋅a⋅d
Isolate d on one side of the equation and solve by plugging your values
d=v2i−v2f2a
d=(15.02−10.02)m2s−22⋅2.0ms−2
d=31.3 m
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
vf=vi−a⋅t, which will get you
t=vi−vfa
t=(15.0−10.0)ms2.0ms2=2.5 s
Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = = = 18.38 rad/sec
Period of oscillation =
= 0.34 sec
I think the correct answer from the choices listed above is option D. One advantage of using electromagnets in devices would be that electromagnets can <span>easily be turned on and off. Hope this answers the question. Have a nice day.</span>
Answer:17.44A
Explanation: Current density=I/Area
Area is given by 2.79mm^2=2.79×10^-6m^2
Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A