Help please ASAP ONLY 1 question

Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.9 cm and a current of 12 A. The bigger lo

Svetradugi [14.3K]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC

OLEGan [10]

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

6 0

3 years ago

Latitude where each of the following types of winds-

Dmitriy789 [7]

westerlies

polar easterlies and trade winds are in a different type

6 0

3 years ago

During a baseball game, a batter hits a popup to a fielder 83 m away. The acceleration of gravity is 9.8 m/s 2 . If the ball rem

topjm [15]

Answer:

41.2 m.

Explanation:

It takes half the time that is (5.8/2) = 2.9 seconds, for the ball to reach its apex.

Given:

S = 83 m

t = 5.8 s

vf = 0 m/s

a = - g

= - 9.81 m/s^2

Equations of motion:

i. vf = vi + a * t

ii. h = vi * t + 1/2 (a * t²)

Using the i. equation of motion:

0 m/s = vi - (9.8 m/s²) (2.9 sec)

vi = 28.4 m/s.

Using the ii. equation of motion:

h = (28.4 * 2.9 ) - 1/2 (9.8 * (2.9)²)

= 82.4 - 41.2

= 41.2 m.

3 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:

$\sum F=ma$