Help me, please

Why do buildings on permafrost have potential structural issues?

Veronika [31]

I believe that the answer to the question provided above is that the <span>buildings on permafrost have potential structural issues because the soil is not stable in this area.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

8 0

3 years ago

Read 2 more answers

An electromagnet is created using a battery, an insulated copper wire and an iron nail. The wire is wrapped around the nail 30 t

Veseljchak [2.6K]

Answer:

Explanation:

If one assume that each turn is like a strand of electromagnet, which can then be added up. Therefore, increase in the number of turns will yield to increase in the magnetic strength. Also if the current increases, then there will be increase in the magnetic field strength.

From Ohm's law

V = IR

I = V/R

That is a direct increase in voltage will lead to increase in current.

Increase the voltage of the battery and increases the number of turns of the coil. Will suit the situation

4 0

2 years ago

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$

let's write this expression for our case

β₁ = 10 log \frac{I_1}{I_o}

β₂ = 10 log \frac{I_2}{I_o}

β₂ -β₁ = 10 ( $log \frac{I_2}{I_o} - log \frac{I_1}{I_o}$ )

β₂ - β₁ = 10 $log \frac{I_2}{I_1}$

log \frac{I_2}{I_1} = $\frac{60 - 20}{10}$ = 3

$\frac{I_2}{I_1}$ = 10³

I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

I = P / A

P = I A

the area is of a sphere

A = 4π r²

the power of the sound does not change, so we can write it for the two points

P = I₁ A₁ = I₂ A₂

I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

I₁ r₁² = (10³ I₁ ) r₂²

r₁² = 10³ r₂²

r₂ = r₁ / √10³

we calculate

r₂ = $\frac{10.0}{\sqrt{10^3} }$

r₂ = 0.316 m

8 0

3 years ago

PLEASE HELP I NEED THIS TO PASS THE EIGHTH GRADE AND I ONLY HAVE A COUPLE HOURS LEFT!!!!!!

zavuch27 [327]

Answer:

the pe at the top of the building: 784 J

the pe halfway through the fall: 392 J

the pe just before hitting the ground: 784 J

Explanation:

Pls brainliest me

I had this question before

7 0

3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago