I need hlep please help a brother out

The diagram below shows a closed system of two tanks that each contain water.

lidiya [134]

Lmalemwlsnlenekenfndelenekf

4 0

2 years ago

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$

The sound intensity of train is 1000 times greater than that of the library.

3 0

2 years ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

3 0

3 years ago

a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and

Cloud [144]

I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx

8 0

3 years ago

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average forc

Natali5045456 [20]

Answer:

Explanation:

Step one:

given data

mass of bullet= 0.02kg

speed v=700m/s

time taken =1.5ms= 0.0015 seconds

Step two:

we know that from the first law

F=ma-----1 first law of motion

also, we know that

a=v/t----2

put a=v/t in equation 1 we have

F=mv/t

Step three:

substitute our given data to find force

F=0.02*700/0.0015

F=14/0.0015

F=9333.33N

F=9.3kN

<u>The average force exerted is 9.3kN</u>

7 0

2 years ago