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3 years ago

A 3.0-kilogram object is placed on a frictionless track at point A and released from rest.

Physics

1 answer:

Eddi Din [679]3 years ago

8 0

Answer:

Explanation:

From the given information:

The gravitational potential energy at A is:

$PE_A = mgh$

GIven that:

mass(m) = 3

height (h) = 3

$PE_A =3\times 9.81 \times 3$

$\mathbf {PE_A =88.29 \ J }$

Using the conservation of mechanical energy:

$PE_A+KE_A = PE_B + KE_B$

$88.29 +0 = 3(9.8)(1) + KE_B$

$88.29 = 29.4 + KE_B$

$KE_B = 88.29 -29.4$

$\mathbf{ KE_B= 58.89 \ J}$

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Explain Newtons Laws

Vsevolod [243]

1) Newton first law of motion: state“If a body is at rest it reamain at rest and if it is in a motion it continu its motion unless a net external force acting on it”

This law also called law of inertia

2) Newton second law of motion: states that ‟acceleration to the force acting on the body and inversky proportional to the mass of the body”

3) Newton's third law of motion: It is not possible to exert a force abody with out the body exerting a forces in the opposite direction.

These forces are called action and reaction forces

hope it's helpful ❤❤❤❤❤❤

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8 0

3 years ago

A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

Nady [450]

Answer:

a) k= 3232.30 N / m, b) f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

w = √k/m

the angular velocity is related to the period

w = 2π / T

we substitute

T = 2 $\pi$ √m/ k

a) empty car

k = 4π² m / T²

k = 4 π² 1310/2 2

k = 12929.18 N / m

This is the equivalent constant of the short springs

F1 + F2 + F3 + F4 = k_eq x

k x + kx + kx + kx = k_eq x

k_eq = 4 k

k = k_eq / 4

k = 12 929.18 / 4

k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

m_total = 1360 + 4 70

m_total = 1640 kg

angular velocity and frequency are related

w = 2pi f

we substitute

2 pi f = Ra K / m

in this case the spring constant changes us

k_eq = 12929.18 N / m

f = 1 / 2π √ 12929.18 / 1640

f = π / 2 2.80778

f = 4,410 Hz

3 0

3 years ago

A clarinet behaves like a tube closed at one end. If its length is 1.0 m, and the velocity of sound is 344 m/s, what is its fund

miv72 [106K]

Answer:

Fundamental frequency = 1376Hz

3 0

4 years ago

A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released

anastassius [24]

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

5 0

3 years ago

The ___________ of a black hole is the radius from a black hole at which the escape velocity is approximately equal to the speed

Vinvika [58]

The Photon Sphere is the radius of the orbit.

3 0

3 years ago