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3 years ago

The track of a roller coaster is shown below. At which point will the riders experience centripetal acceleration?

Physics

2 answers:

bulgar [2K]3 years ago

8 0

The answer to this question is Y

Hope this helps :)

valentinak56 [21]3 years ago

4 0

Centripetal acceleration is applied in circular motion. Y is the point on the track that most closely resembles a circular path for the roller coaster.

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Baseball Digest is but one of several publications on the sport of baseball. True or False?

Gnoma [55]

Answertrue

Explanation:

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3 years ago

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At which angle must a laser beam enter the water for no refraction to occur?

abruzzese [7]

Light that enters the new medium perpendicular to the surface keeps sailing straight through the new medium unrefracted (in the same direction).

Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story: If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it straight on to the surface, perpendicular to it.

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3 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

3 years ago

A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0°. Once it has been set into

Bas_tet [7]

Answer:

The answer is....i am sorry idk

8 0

3 years ago

A 10kg oil is heated by a source of heat that supplies 800J of heat every minute. Calculate The heat added to oil after 30 minut

goblinko [34]

Answer:

4,200 joules per kilogram per degree Celsius

Explanation:

The specific heat capacity of a material is the energy required to raise one kilogram (kg) of the material by one degree Celsius (°C). The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

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2 years ago