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Alex777 [14]
3 years ago
10

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block?
A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Physics
1 answer:
nignag [31]3 years ago
8 0

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block.  The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that  

[tex]f=mgsin \theta [tex]

Therefore, option B is the correct option.

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Agata [3.3K]


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Astronomers announced in August 2016 that they may have found such a planet orbiting Proxima Centauri. The newfound world, known as Proxima b, is about 1.3 times more massive than Earth, which suggests that the exoplanet is a rocky world, researchers said. The planet is also in the star's habitable zone, just 4.7 million miles (7.5 million kilometers) from its host star. It completes one orbit every 11.2 Earth-days. As a result, it's likely that the exoplanet is tidally locked, meaning it always shows the same face to its host star, just as the moon shows only one face (the near side) to Earth.



The young sun would have had a very strong magnetic field, whose lines of force reached out into the disk of swirling gas from which the planets would form. These field lines connected with the charged particles in the gas, and acted like anchors, slowing down the spin of the forming sun and spinning up the gas that would eventually turn into the planets. Most stars like the sun rotate slowly, so astronomers inferred that the same “magnetic braking” occurred for them, meaning that planet formation must have occurred for them. The implication: Planets must be common around sun-like
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Most of the first exoplanet discoveries were huge Jupiter-size (or larger) gas giants orbiting close to their parent stars. That's because astronomers were relying on the radial velocity technique, which measures how much a star “wobbles” when a planet or planets orbit it. These large planets close in produce a correspondingly big effect on their parent star, causing an easier-to-detect wobble.
Before the era of exoplanet discoveries, instruments could only measure stellar motions down to a kilometer per second, too imprecise to detect a wobble due to a planet. Now, some instruments can measure velocities as low as a centimeter per second, according to Matthews. “Partly due to better instrumentation, but also because astronomers are now more experienced in teasing subtle signals out of the data.”

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Kepler has revealed a cornucopia of different types of planets. Besides gas giants and terrestrial planets, it has helped define a whole new class known as “super-Earths”: planets that are between the size of Earth and Neptune. Some of these are in the habitable zones of their stars, but astrobiologists are going back to the drawing board to consider how life might develop on such worlds.

In 2014, Kepler astronomers (including Matthews’ former student Jason Rowe) unveiled a “verification by multiplicity” method that should increase the rate at which astronomers promote candidate planets to confirmed planets. The technique is based on orbital stability — many transits of a star occurring with short periods can only be due to planets in small orbits, since multiply eclipsing stars that might mimic would gravitationally eject each other from the system in just a few million years.

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6 0
4 years ago
Read 2 more answers
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
PLSSSSS HELP!!!!!!! ASAP!!!!! MARK BRAINLIEST!!!!
Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

5 0
3 years ago
He has 54 cds and every box has 9 cds.how many boxes are in 54 cds?
Gemiola [76]

Solution:
54 / 9 = 6 boxes.
6 0
3 years ago
From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those t
Dominik [7]

'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.

'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.

'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.

<u>Explanation:</u>

Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.

As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.

So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.

And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.

5 0
3 years ago
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