The diagram represents water waves travelling from deep water into an area of much shallower water.

natulia [17]

Entropy is an extensive property of a thermodynamic system. It quantifies the number Ω of microscopic configurations (known as microstates) that are consistent with the macroscopic quantities that characterize the system (such as its volume, pressure and temperature).[1] Under the assumption that each microstate is equally probable, the entropy
S
S is the natural logarithm of the number of microstates, multiplied by the Boltzmann constant

8 0

3 years ago

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a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v₁.

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v₁.

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t

erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0

4 years ago

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Please anyone post any class 9 physics question

aalyn [17]

Answer:

I will but can you just wait for some minutes cus I am in a hurry now.

sorry that pic is a little blurry

6 0

3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0

4 years ago