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katrin [286]
3 years ago
7

(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a

Physics
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

t=0.417s

Explanation:

After the ball hits the racket it is in freefall(assume air resistance as negligible)

so a=-g

use

x-x0=v0t+1/2at^2

Plug in givens

20=50t-4.9t^2

Solve quadratic equation using quadratic formula

t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)

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Forensic ______________ focuses on the relationship between human behavior and criminal justice.
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Anthropology. It focuses on human behavior. 
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EASY MATCHING PLEASE HELP!
kakasveta [241]
<h2><u>Answers:</u></h2><h2>1.) Right answer: polarization  </h2>

Alignment of light in only one vibrational orientation: polarization  

Normally, electromagnetic waves (light) are not polarized, so electromagnetic vibration occurs in all planes. But when we get these waves to vibrate in a single plane, we have polarized light.

This is possible because electromagnetic waves are transversal waves, this means the electric field oscillates in all normal directions to the direction of wave propagation.

In other words:

<em>Its oscillation occurs in the transversal direction to its propagation. </em>

So, when polarization occurs the electric field vibrates or oscillates in a given plane, called the <u>polarization plane</u>.

It should be noted that this phenomenon is only possible in transversal waves, in longitudinal waves, such as sound waves, polarization is not possible because its oscillation occurs in the same direction as its propagation.

<h2>2.) Right answer: concave  </h2>

Mirror that causes parallel incident rays of light to converge at the focus: concave  

A concave mirror, or convergent mirror, has a reflective surface that is curved inwards.

The concave mirrors reflect the light making it converge in a focal point therefore they are used to focus the light. This occurs because the light is reflected with different angles, since the normal to the surface varies from one point to another of the mirror.

<h2>3.) Right answer: photoelectric effect </h2>

phenomenon that can be explained only by a particle model for light: photoelectric effect  

Light can be considered as a wave or as particles (photons), in this context the photoelectric effect can only be explained based on the corpuscular model (particles) of light.

Then, the photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If the light is a stream of photons and each of them has energy, this energy is be able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

 

<h2>4.) Right answer: Taylor's experiment   </h2>

experiment that showed that diffraction effects could be attributed to light particles: Taylor's experiment  

This experiment was carried out by Geoffrey Taylor in 1909 with a flame as a light source, a diffraction grating and a photographic plate.

All this to test the diffraction of light.


<h2> 5.) Right answer: reflection  </h2>

principal use of mirrors: reflection

Mirrors fulfill the principle of reflection, which occurs when the light rays fall on a very flat reflecting surface are reflected so that the incident angle is equal to the reflected angle

<h2>6.) Right answer: mirage  </h2>

image of the sky seen on a hot road: mirage

A mirage is the product of an optical illusion due to the total reflection of the light when crossing layers of hot air of different density; this causes the perception of the inverted image of distant objects, as if they were reflected in the water.

<h2>7.) Right answer: virtual  </h2>

type of images always produced by convex  mirrors: virtual

In the convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object.

<h2>8.) Right answer: diffraction  </h2>

pattern produced by light through a narrow slit: diffraction

Diffraction is a phenomenon that is based on the deviation of the waves (light waves in this case) when encountering an obstacle or going through a slit

<h2>9.) Right answer: convex  </h2>

shape of a converging lens: convex  

A convex lens is thicker in the center than at its edges and concentrates (converges) at a point the rays of light that pass through it.

<h2>10.) Right answer: dispersion  </h2>

separating light into component colors: dispersion  

The dispersion of light occurs when a beam of composite light is refracted (the different rays of light are diverted depending on their frequencies) in some medium, leaving their constituent colors separated.

The best known case is when a beam of white light from the sun passes through a prism, thus obtaining rays of different colors like those of the rainbow.


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How many neutrons does element X have if its atomic number is 23 and its mass number is 90?
givi [52]

Answer:

67

Explanation:

- The atomic number (Z) of an atom is equal to the number of protons in the nucleus

- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus

Therefore, calling p the number of protons and n the number of neutrons, for element X we have:

Z = p = 23

A = p + n = 90

Substituting p=23 into the second equation, we find the number of neutrons:

n = 90 - p = 90 - 23 = 67

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A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
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