Everyone is trying to living a perfect life via social media.

I don’t know what to do help me

kirill115 [55]

Answer:

oooh thats hard

Explanation:

well i would probaly search the page number or anthing on the page that should help

4 0

2 years ago

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

Vesna [10]

Answer:

Number of turns per unit length will be $n=2.786\times 10^4turns/m$

Explanation:

We have given that strength of the magnetic field produced by the solenoid B = 7 T

Current in the solenoid i = 200 A

Let the number of turns per unit length is n

Magnetic field due to solenoid is given as $B=\mu ni$ here $\mu$ is permeability of free space n is number of turns per unit length and i is current

So $7=4\pi \times 10^{-7}\times n\times 200$

$n=2.786\times 10^4turns/m$

7 0

2 years ago

Read 2 more answers

Help with this question please.

mr Goodwill [35]

There are 4 hydrogens on the right side $(2\mathrm H_2=4\mathrm H)$ , and 2 hydrogens on the left per molecule of $\mathrm H_2$ . To get the same number of hydrogens on both sides, the coefficient should be 2.

(Then the number of oxygens will be consistent, since $2\mathrm H_2\mathrm O$ contributes 2 oxygens, and so does $\mathrm O_2$ .)

5 0

3 years ago

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest

mr_godi [17]

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

$\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$