2. Which are most closely associated with an element on the periodic table?

Convert 4.565mm to m

lutik1710 [3]

It will be 0.004565 m

4 0

4 years ago

Why does the output of a microphone increase as the frequency of the sound waves which it receives increases

Sloan [31]

Answer:

See Explanation

Explanation:

The frequency of sound waves received by the microphone influences the output or pitch of the sound obtained from the microphone.

The higher the frequency of the sound received by the microphone, the higher the output of the microphone and vice versa. This is because, the higher the frequency of sound, the higher the oscillations produced and the greater the output of the microphone.

The rise and fall in the pitch of sound waves as the frequency of sound waves varies is called inflection.

7 0

3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Nezavi [6.7K]

Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=2.80 m/s$ is the initial speed

$\theta=39\°$ is the angle at which the venom was shot

$t$ is the time since the venom is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.4 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin:

First we have to find $t$ from (2):

$0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}$ (3)

Rearranging (3):

$-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0$ (4)

This is a <u>quadratic equation</u> (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9$

$b=1.762$

$c=0.4$

Substituting the known values:

$t=\frac{-1.762 \pm \sqrt{(1.762)^{2} - 4(-4.9)(0.4)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.517 s$ (7)

Substituting (7) in (1):

$x=2.8 m/s cos(39\°) (0.517 s)$ (8)

Finally:

$x=1.124 m$ (9)

4 0

4 years ago

Find the force exerted by each rope in the sign below

mixas84 [53]

Answer: Example 1: Consider a crate being pulled along a frictionless floor (while such a floor is very hard to find, this will still help us understand the concept and we can return to this situation later, after considering friction, and solve it more realistically).

Consider a crate being pulled along a horizontal, frictionless floor. A rope is tied around it and a man pulls on the rope with a force of T. T is the tension in the rope. What happens to the crate?

Before we can apply Newton's Second Law,

F = m a

we must find the net force -- the vector sum of all the forces -- acting on the object. In addition to the force T exerted by the rope, what other forces act on the object?

As discussed in class, in Mechanics, we can restrict our attention to "contact" forces and "gravity". That means gravity pulls down on this crate with a force equal to its weight, w. But the floor supports the crate. The floor responds by pushing up on the crate with a force we call the normal force. "Normal" means "perpendicular". We will call this force n; you may also encounter it labeled N or FN.

Explanation:

4 0

3 years ago

A positively charged particle initially at rest on the ground accelerates upward to 200m/s in 2.60s . The particle has a charge-

Butoxors [25]

Answer:

E = 867 N/C, upward.

Explanation:

Assuming no other forces acting on the particle, it must obey Newton's 2nd Law, as follows:

$F_{net} = m*a (1)$

There are two forces acting on the particle in the vertical direction, one due to the electric field, and the other due to gravity.
Since the particle is positively charged, assuming that the electric field aims upward, we can write the following expression for the left side of (1):

$F_{net} = q*E - m*g = m*a (2)$

Since we know that q/m = 0.1 C/kg,
⇒ q = 0.1m C/kg
Replacing this value of q in (2), and simplifying masses, we get:

$F_{net} = 0.1 *E - 9.8 m/s2 = a (3)$

We can find the value of a, simply applying the definition of acceleration:

$a =\frac{\Delta v}{\Delta t} = \frac{200m/s}{2.6s} =76.9 m/s2 (4)$

Replacing (4) in (3) and solving for E, we get:
E = 867.2 N/C, upward.

6 0

3 years ago