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3 years ago

What is the number of wavelengths that pass a given point each second called?

1 answer:

4 0

Since waves are moving, we define frequency as the number of waves that pass a given point in a specified unit of time. The unit commonly used is Hertz which is the number of wave cycles that pass a point in one second.

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A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. The correct form is: A $+10\,\mu C$ charge and a $-10\,\mu C$ at a distance of 0.3 meters.

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$

$F = 9.987\,N$

In consequence, correct answer is B.

4 0

3 years ago

If the vertical component of velocity for a projectile is 7.3 meters/second, what is its hang time?

Kamila [148]

A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.

So, the hang up time, or the time the object stayed on air is calculated using this equation:

a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds

6 0

3 years ago

Is it possible for an object to have a charge of 4.8x10-9 C? Why or why not?

Gelneren [198K]

I have absolutely no clue

3 0

3 years ago

Most of uranus's atmosphere is composed of

Vladimir [108]

So the question is what gases form the atmosphere of Uranus. So Uranus and Neptune are classified as "ice giants". They have a similar atmosphere to "gas giants" Saturn and jupiter and Neptunes atmosphere is primarily composed out of hidrogen and helyum. So the correct answer is a.

5 0

3 years ago

Read 2 more answers

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

2 years ago