All categories

3 years ago

Describe a situation in which jad would be not be ideal.

1 answer:

5 0

Students ought to depict one of the accompanying circumstances:
(1) User gatherings are eager and need something new, not a standard answer for a run of the mill issue,
(2) The hierarchical culture bolsters joint critical thinking practices among various levels of representatives,
(3) Analysts estimate that the quantity thoughts produced through one-on-one meetings won't be as copious as the quantity thoughts conceivable from an expanded gathering exercise, or
(4) Organizational work process allows the nonattendance of key faculty amid a two-to-four-day piece of time.

You might be interested in

What mass of a material with density ρ is required to make a hollow spherical shell having inner radius r1 and outer radius r2?

krok68 [10]

Answer: $m=\frac {\rho}{\frac{4}{3} \pi( (r_2)^{3} - (r_1)^{3}) }$

It is given that:

Density of material =ρ

radius of inner spherical shell = $r_1$

radius of outer spherical shell= $r_2$

we know that the volume of sphere is $\frac{4}{3}\pi r^{3}$

Volume of the given spherical shell = $\frac{4}{3}\pi( (r_2)^{3}-(r_1)^{3})$

Then mass of the spherical shell can be calculate as:

Mass, m=density/ volume

$m=\frac {\rho}{\frac{4}{3} \pi ((r_2)^{3} - (r_1)^{3}) }$

6 0

3 years ago

True or false? "The internal energy is the total energy stored in the bonds of a sample."

kotykmax [81]

trueExplanation:becuse true ihuorhileklduowelkhds

4 0

3 years ago

Uuse Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire loop. You will need to u

Alex777 [14]

Answer:

Explanation:

According to the Fleming's right hand rule, if we spread our right hand such that the thumb, fore finger and the middle finger are mutually perpendicular to each other, then the thumb indicates the direction of force, fore finger indicates the direction of magnetic field, then the middle finger indicates the direction of induced current.

According to the Lenz's law, the direction of induced emf is such that it always opposes the cause due to which it is produced.

4 0

3 years ago

The IMA is always smaller than the AMA.<br><br> True<br> False

VMariaS [17]

Best answer must be false. Because the IMA it doesn't have always stay smaller than the AMA. Hope it helped you!

6 0

4 years ago

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$