An electric current will always follow

When a 4.25 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.62 cm. what is the force cons

ANTONII [103]

The force applied to the spring is the weight of the object that compresses it, so it is equal to:
$W=mg=(4.25 kg)(9.81 m/s^2)=41.7 N$

Because of this force, the spring compresses by $x=2.62 cm=0.0262 m$ . Using Hook's law,
$F=kx$ ,
since we know the intensity of the force (the weight W) and the compression of the spring, x, we can find k, the spring constant:
$k= \frac{F}{x}= \frac{W}{x}= \frac{41.7 N}{0.0262 m} =1591.6 N/m$

6 0

3 years ago

A particle of unit mass moves in a potential V(x)= ax^2+b/x^2 . Where a and b are constnts. Calcuate the angular frequency of sm

Sholpan [36]

Answer:

Explanation:

Given

Potential Energy is given by

$U(x)=ax^2+\frac{b}{x^2}$

And Force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

Particle will be at equilibrium when Potential Energy is either minimum or maximum

$F=-\left ( 2ax-\frac{2b}{x^3}\right )$

i.e. $ax=\frac{b}{x^3}$

$x_0=(\frac{b}{a})^{0.25}$

So angular Frequency of small oscillation is given by

$\omega =\sqrt{\frac{U''(x)}{m}}$

for $m=1$

we get $\omega =\sqrt{\frac{U''(x_0)}{1}}$

$U''(x_0)=2a+6a= 8a$

$\omega =\sqrt{8a}$

4 0

3 years ago

You walk 20 meters north, then 5 meters south. what is your displacement?

Svetllana [295]

The answer is C. 15 meters North

Displacement is the distance and direction from the origin. If you move 20 meters North, your displacement is 20 meters North. If you move 5 meters South, you are basically moving back the way you came by 5 meters, which means you are now 15 meters North.

I hope this helped! :)

3 0

3 years ago

Time Warner Cable's leadership development program that spanned over 30 days and included weekly videos, practice exercises, and

ankoles [38]

Answer: A.

tracking training through a leaming records store LRS.

Explanation:

An LRS uses xAPI to collect learner data, or experiences, from both online and offline sources. These experiences are reported back to the LRS in the form of xAPI statements, where they are stored. These statements can then be retrieved for reporting and interpretation of the learner data.

8 0

3 years ago

A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof mak

nekit [7.7K]

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

The horizontal speed
the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

$v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s$

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

$x(t)=8.8 m/s *t$

$y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}$

Now we calculate the time the hammer takes to get to the floor

$17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s$ or $t=-2.38s$

Now, we keep the positive time result and calculate the horizontal distance travelled

$x(1.47s)=8.8m/s*1.47s=12.94m$

3 0

3 years ago