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stellarik [79]
3 years ago
14

Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker

B immediately accelerates when he is passed, catching up with biker A after 1.00 minute.
(a) Draw graphs of position, velocity, and acceleration of both bikers as a function of time.
(b) What is the acceleration of biker B, assuming that it is constant?
(c) At what distance from the origin do the two bikers meet?
(d) What is the average speed of biker B during the first 30.0 seconds?

Physics
1 answer:
Burka [1]3 years ago
3 0

Answer: attached

Explanation:

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Explanation:

<h2><u>Steps </u><u>:</u></h2>
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7 0
2 years ago
A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

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3 years ago
Which is a molecule? Select one: a. graphene b. NaCl c. Ne d. trail mix
Ede4ka [16]
What Is A Molecule? B) NaCi
8 0
3 years ago
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Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to
Zigmanuir [339]

Answer:

gamma rays , X rays,  ultraviolet , visible light , infrared,  radio waves

Explanation:

The electromagnetic spectrum is the set of electromagnetic radiations distributed in their different frequencies or wavelengths, which in turn are related to their energy.   If we go from the smallest wavelengths known up to now (because according to physics the electromagnetic spectrum is infinite and continuous) to the longest, the electromagnetic spectrum covers the following radiations:  

Gamma rays, X-rays, ultraviolet, visible light (all the colors we are able to see), infrared, radio waves and microwaves.  

Let's make a brief of them:

-Gamma rays: With a wavelength in the order of 10^{-12}m, is a type of ionizing radiation capable of penetrating matter quite deeply and is able to cause serious damage to the nucleus of the cells. Inaddiito, these rays are used to sterilize medical equipment and food.

-X rays: With a wavelength between 1m and 10km. It is invisible to the human eye, capable of crossing opaque bodies and of being an ionizing radiation.

-Ultraviolet light: Whose wavelength is approximately between 100 nm and 380 nm; is a type of electromagnetic radiation that is not visible to the human eye.

-Visible light: This part of the spectrum is located between ultraviolet light and infrared light (380 nm - 780 nm).  It should be noted, the fact the only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

-Infrared: This type of radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm).  

These waves can be divided into:  

<u>- Near infrared</u> or long wave infrared: it is the least sensitive to color and is easily absorbed by water.  

<u>- Medium or medium wave infrared:</u> it is also insensitive to color and easily absorbed by water and many types of plastics and paints.  

-<u> Far infrared or short wave infrared: </u>it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.  

This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

-Radio waves: These are a type of electromagnetic radiation with wavelengths between 10 m to 10,000 m. This type of electromagnetic waves is very well reflected in the ionosphere, the layer of the atmosphere through which they travel directly or using repeaters.  In addition, they are very useful to transport information, being important in telecommunications. They are used not only for conventional radio transmissions but also in mobile telephony and TV.  

5 0
3 years ago
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