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3 years ago

8

A compressed spring is compressed between a 5kg and 3 kg cart, respectively. When the spring is released, what will be the same

for the carts?

1 answer:

stira [4]3 years ago

4 0

The momentum of the two carts will be the same

Explanation:

First of all, we notice that the system consisting of the two carts + the spring is an isolated system: this means that there are no external forces acting on it.

As a result, the total momentum of the system must be conserved.

Since the spring is at rest, the total momentum of the system after the two carts are released is:

$p_f = p_1 + p_2$

where

$p_1$ is the momentum of the 1st cart

$p_2$ is the momentum of the 2nd cart

The total momentum of the system at the beginning, when the spring is compressed, is zero:

$p_i=0$

Since the total momentum must be conserved, we have:

$p_i = p_f\\0 = p_1 + p_2\\\rightarrow p_2 = -p_1$

Which means that the two carts will have equal and opposite momenta.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

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Light from a fluorescent lamp is observed through a cloud of cool nitrogen gas. Again, two students are having a discussion abou

Vadim26 [7]

Answer:

From the previous explanation Student No. 1 has the correct explanation

Explanation:

When the fluorescent lamp emits a light it has the shape of its emission spectrum, this light collides with the atoms of Nitrogen and excites it, so these wavelengths disappear, lacking in the spectrum seen by the observed, for which we would see an absorption spectrum

The nitrogen that was exited after a short time is given away in its emission lines, in general there are many lines, so the excitation energy is divided between the different emission lines, which must be weak

From the previous explanation Student No. 1 has the correct explanation

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3 years ago

If K= -3 and n= 9 what is -5+-3-9

kherson [118]

Answer:

-17

Explanation:

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3 years ago

What is the most potentially destructive way to deal with stress while at college? a. texting friends back home b. watching TV c

diamong [38]

Answer:

Option (C)

Explanation:

Recreational drugs are usually referred to those drugs or medicines that are legally or illegally used without any prescription given by the doctors. These drugs include marijuana, cocaine, alcohol, nicotine, heroin, and many others. These drugs provide relaxation for a shorter period of time, but using of these substances, it makes an individual addicted to drugs.

So, recreational drugs are one of the most potentially destructive ways to handle stress when a student is in college.

Thus, the correct answer is option (C).

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3 years ago

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

or

$\frac{Q_2}{3.78\times 10^3}=0.72$

or

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

or

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

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3 years ago

A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450

Natasha2012 [34]

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$ .

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$

The initial speed of the bullet-block system is:

$v \approx 2.383\,\frac{m}{s}$

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

$(0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)$

$v_{o} = 889.199\,\frac{m}{s}$

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3 years ago