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3 years ago

What is the equation of instantaneous velocity

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

7 0

Answer:

it's possible to calculate an object's velocity at any moment along its path. This is called instantaneous velocity and it is defined by the equation v = (ds)/(dt),in other words, the derivative of the object's average velocity equation.

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The smallest possible part of an element that can still be identified as the element is called

andreyandreev [35.5K]

Answer:

BLM

Explanation:

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3 years ago

a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height

ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′ and mass ′

M′ is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

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3 years ago

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3 years ago

Read 2 more answers

Each of the gears a and b has a mass of 675 g and has a radius of gyration of 40 mm, while gear c has a mass of 3. 6 kg and a ra

navik [9.2K]

9.87 seconds

The time required for this system to come to rest is equal to 9.87 seconds.

We have the following data:

Mass of gear A = 675 g to kg = 0.675 kg.

Radius of gear A = 40 mm to m = 0.04 m.

Mass of gear C = 3.6 kg.

Radius of gear C = 100 mm to m = 0.1 m.

How can I calculate the time needed?

We would need to figure out the moment of inertia for gears A and C in order to compute the time needed for this system to come to rest.

Mathematically, the following formula can be used to determine the moment of inertia for a gear:

I = mr²

Where:

m is the mass.

r is the radius.

We have, For gear A:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

We have, For gear C:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

The initial angular velocity of gear C would therefore be converted as follows from rotations per minute (rpm) to radians per second (rad/s):

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

On Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549 ----->equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549 ------>equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098 ------>equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398 ------->equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

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2 years ago

Please help I don't know how to answer these questions!

Yuki888 [10]

1) The potential energy is the most at the highest position and the least at the equilibrium position

2) The kinetic energy is the most at the equilibrium position and the least at the highest position

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:

$PE\propto h$

This means that:

The potential energy is the most when the bob is at the highest position
The potential energy is the least when the bob is at the equilibrium position, which is the lowest position

We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,

$E=KE+PE=const.$

where KE is the kinetic energy.

From the equation above, we observe that:

When PE is maximum, KE must be at minimum
When PE is minimum, KE must be maximum

Therefore, this implies that:

The kinetic energy is the most when the potential energy is the least, i.e. at the equilibrium position
The kinetic energy is the least when the potential energy is the most, i.e. at the highest position

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3 years ago

What is the equation of instantaneous velocity​

What is the equation of instantaneous velocity