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mr_godi [17]
3 years ago
9

How do you calculate centripetal acceleration?

Physics
1 answer:
Delvig [45]3 years ago
8 0

A particle has centripetal acceleration whenever it's a making a turn of radius R. If the particle is moving at a constant tangential speed v throughout the turn, then the magnitude of centripetal acceleration is

v²/R

If the particle is following a uniformly circular path, then it moves in a circle of radius R and travels a distance equal to its circumference, 2πR. Let T be the time it takes to complete one such loop. Then the entire circle is traversed with speed v = 2πR/T, so that the centripetal acceleration is also given by

v²/R = (2πR/T)²/R = 4π²R/T²

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the diagram shows a ray of light striking a mirror that is above a sink full of water explain what what will happen to the beam
ratelena [41]
The water will reflect some of the light
6 0
4 years ago
Read 2 more answers
2<br><br> How does Descartes' "quality of motion" differ from the modern<br><br> momentum?
Rzqust [24]

Answer: Descartes was more of speed which defers from modern day velocity.

Explanation:

Descartes law if conservation referred or defined “motion” rather than “momentum” as what is obtainable in today's world as ”speed” the rate at which something moves rather than “velocity” which is a product of speed and direction. So in conclusion Descartes was more of speed which defers from modern day velocity.

6 0
3 years ago
Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the
True [87]

Answer:

E=54V/cm

\Delta V=496.8V between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

\Delta V=Ed

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

\Delta V=Ed=(54V/cm)(9.2cm)=496.8V

8 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal forces, what is the wagon
s2008m [1.1K]

Answer:

The acceleration of the wagon is 3 m/s².

To calculate the acceleration of the wagon, we use the formula below.

Formula:

F = ma............. Equation 1

Where:

F = horizontal Force

m = mass of the wagon

a = acceleration of the wagon.

make a the subject of the equation

a = F/m.............. Equation 2

From the question,

Given:

F = 30 N

m = 10 kg

Substitute these values into equation 2

a = 30/10

a = 3 m/s²

Hence, the acceleration of the wagon is 3 m/s².

5 0
2 years ago
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