PHYSICS HIGH SCHOOL

A cheetah runs with a positive acceleration. What does this tell you about the velocity time graph for the cheetah?A. Graph negative slope
B. graph positive slope
C. Graph positive area
D. Graph negative area

Answers

Answer 1
Answer: B is the correct answer
Answer 2
Answer:

Answer:

B. graph positive slope

Explanation:

got it right

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Deimos completes one (circular) orbit of Mars in 1.26 days. The distance from Mars to Deimos is 2.35×107m. What is the centripetal acceleration of Deimos?

Answers

Answer:

The centripetal acceleration of Deimos is 0.077m/s^(2).

Explanation:

The centripetal acceleration is defined as:

a = (v^(2))/(r)   (1)

Where v is the velocity of Deimos and r is the orbital distance.        

Notice that is necessary to determine the velocity first.

The speed of the Deimos can be found by means of the Universal law of gravity:

F = G(M \cdot m)/(r^(2))  (2)

Then, replacing Newton's second law in equation 2 it is gotten:

m\cdot a = G(M \cdot m)/(r^(2))  (3)

However, a is the centripetal acceleration since Deimos almost describes a circular motion around Mars:

a = (v^(2))/(r)  (4)

Replacing equation 4 in equation 3 it is gotten:

m(v^(2))/(r) = G(M \cdot m)/(r^(2))

m \cdot v^(2) = G (M \cdot m)/(r^(2))r

m \cdot v^(2) = G (M \cdot m)/(r)

v^(2) = G (M \cdot m)/(rm)

v^(2) = G (M)/(r)

v = \sqrt{(G M)/(r)}  (5)

Where v is the orbital speed, G is the gravitational constant, M is the mass of Mars, and r is the orbital radius.    

v = \sqrt{((6.67x10^(-11)N.m^(2)/kg^(2))(6.39x10^(23)kg))/(2.35x10^(7)m)}

v = 1346m/s

Finally, equation 4 can be used:

a = ((1346m/s)^(2))/(2.35x10^(7)m)

a = 0.077m/s^(2)

Hence, the centripetal acceleration of Deimos is  0.077m/s^(2).

Final answer:

The centripetal acceleration of Deimos, one of Mars' moon, can be calculated using its orbital period and distance from Mars. Convert the time units to seconds and use the formulas for velocity and centripetal acceleration to get an answer of approximately 7.84x10^-5 m/s^2.

Explanation:

To find the centripetal acceleration of Deimos, we can use the formula for centripetal acceleration, which is a =v^2/r , where v is the velocity and r is the radius (distance from Mars to Deimos). The velocity can be found using the formula v = 2πr/T, where T is the period (time for one complete orbit).

First, convert the days into seconds because the SI unit of time in physics is second. So, 1.26 days = 1.26 * 24 * 60 * 60 = 108864 seconds.

Then, calculate the velocity: v = 2 * π * 2.35x10^7m / 108864s = 1.36 km/s.

Finally, substitute v and r into the centripetal acceleration formula: a = (1.36x10^3m/s)^2 / 2.35x10^7m = 7.84x10^-5 m/s^2.

The centripetal acceleration of Deimos is approximately 7.84x10^-5 m/s^2.

Learn more about Centripetal Acceleration here:

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According to the information in your study unit, which of the following locations requires the highest illuminance? A. A residential stairway
B. A school locker room
C. A bank lobby
D. A study area

Answers

The correct answer is letter D. A study area. According to the information in your study unit, the location that requires the highest illuminance is the study room because it is the place the students can stay to study. The more brighter the place is, the more people are using that area.

Which of the following is true about light with a single wavelength?1) It can be refracted
II) It cannot be dispersed
III) It can be reflected *

Answers

Answer:

It can be refracted

Explanation:

When a single wavelength wave which is known as a monochromatic light is passed from one transparent medium to the other, it can refract. This refraction can only be seen if the light enter the medium at some angle.

A glass lens, n glass = 1.52 , has a 131 nm thick antireflective film coating one side, n film = 1.25. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating?

Answers

Answer:

655 nm

Explanation:

When the intereference is destructive then the thickness, d of antireflective film coating one side is given by

2d=w/2n

Where w is wavelength and n is the reflective index of the film

Making w the subject of formula then

w=4nd

Substituting 1.25 for n and 131 nm for d then the wavelength will be

w=4*1.25*131=655 nm

Therefore, the wavelength is equivalent to 655 nm

The formula for calculating the wavelength in an antireflective film involves thickness (d) and refractive index (n). For n = 1.25 and d = 131 nm, the resulting wavelength is 655 nm.

When light waves encounter a thin film, some of the waves are reflected from the top surface of the film, and some pass through it. These waves can interfere with each other, leading to constructive or destructive interference. In the case of antireflective coatings, destructive interference is desired to minimize reflection.

The formula you mentioned is used to calculate the thickness (d) of an antireflective film that results in destructive interference for a specific wavelength (w) of light. The formula is:

2d = w / (2n)

Where:

d is the thickness of the film.

w is the wavelength of light.

n is the refractive index of the film.

To find the wavelength (w) when given the thickness (d) and refractive index (n), you can rearrange the formula:

w = 4 * n * d

Now, let's calculate the wavelength using the provided values:

n = 1.25 (refractive index)

d = 131 nm (thickness in nanometers)

Substitute these values into the formula:

w = 4 * 1.25 * 131 = 655 nm

Therefore, the calculated wavelength (w) is 655 nanometers (nm). This means that for a film with a refractive index of 1.25 and a thickness of 131 nm, destructive interference occurs at a wavelength of 655 nm.

For more such information on: wavelength

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Four landing sites are proposed for a lander. Data about the sites are listed in the table.A 3 column table with 4 rows. The first column is labeled landing site with entries W, X, Y, Z. The second column is labeled height above surface in meters with entries 32, 16, 35, 12. The last column is labeled acceleration due to gravity in meters per second squared with entries 1.6, 3.7, 1.6, 3.7.
At which landing site would the lander have the greatest amount of gravitational potential energy?

A. W
B. X
C. Y
D. Z

Answers

Answer:

B. X.

Explanation:

To determine which landing site would have the greatest amount of gravitational potential energy, we need to consider the height above the surface and the acceleration due to gravity at each site.

Gravitational potential energy is given by the formula:

Gravitational potential energy = mass x acceleration due to gravity x height

In this case, the mass of the lander is not provided, but since it is the same for all the sites, we can ignore it for the purpose of comparison. Therefore, we only need to consider the acceleration due to gravity and the height above the surface.

Looking at the table, we can see that at site X, the height above the surface is 16 meters, and the acceleration due to gravity is 3.7 meters per second squared. This means that at site X, the lander would have the highest amount of gravitational potential energy compared to the other sites.

i hoped this helped !   ⚫w⚫

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positive direction, what are John and Betty's speeds?

Answers

Answer:

The John 's speed is 20 m/s.

The Betty's speed is 20 m/s

Explanation:

Given that,

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second.

We need to calculate the John and Betty's speeds

We know that,

The speed is a scalar quantity. its has only magnitude.

Hence, The John 's speed is 20 m/s.

The Betty's speed is 20 m/s
Both of their speeds are 20 m/s.
Direction is not involved in describing speed.
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