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3 years ago

Chemical properties of an acid include...

1 answer:

5 0

and C. corrosive, increases the concentration of hydrogen ions when added to water, forms hydrogen gas when it comes in contact with a metal, and formssalt and water when added to a base.

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A particle is acted on by two torques about the origin: τ→1 has a magnitude of 8 N·m and is directed in the positive direction o

lyudmila [28]

To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.

For the given problem we have to

$T_i = 8Nm$

$T_j = -8.9Nm$

In this way the torque acting on the particle as a function of distance and time is,

$\tau = \frac{dL}{dt} = 8\hat{i}-8.9\hat{j}$

The net torque acting on the particle is

$\tau_{net} = \sqrt{T_i^2+T_j^2}$

$\tau_{net} = \sqrt{(8)^2+(-8.9)^2}$

$\tau_{net} = 11.967Nm$

PART B) The direction of the torque is given by,

$tan\theta = \frac{y}{x}$

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}(\frac{-8.9}{8})$

$\theta = -48.04\°$

Therefore the torque direction is 48.04° below the x axis.

5 0

4 years ago

At a depth of about 10 meters the water pressure on you is equal to the pressure of 1 atmosphere. the total pressure on you is t

MrRa [10]

It would be 22 depths inside the glass of water.

8 0

4 years ago

Help me with this please

NARA [144]

I think 9 is function

5 0

3 years ago

Read 2 more answers

A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a

sladkih [1.3K]

Answer:

$d=1.29*10^{-6}m$

Explanation:

From the question we are told that:

Distance of wall from CD $D=1.4$

Second bright fringe $y_2= 0.803 m$

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

$y=frac{n*\lambda*D}{d}$

Where

$d=\frac{n*\lambda*D}{y}$

$d=\frac{2*431 *10^{-9}m*1.4}{0.803}$

$d=1.29*10^{-6}m$

8 0

3 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

<u>Where:</u>

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

4 years ago