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3 years ago

Una locomotiva traina un solo vagone, che ha la sua stessa massa, con la forza di

Physics

1 answer:

gregori [183]3 years ago

3 0

Answer:

c) F = 16000 N

Explanation:

For this exercise we use Newton's second law

F = ma

they tell us that adding the other wagons the acceleration of the locomotive must be maintained

F = m a

by adding the other four wagons

mass = 4 no

therefore to maintain the force you must also raise the same factor

Fe = 4Fo

Fe = 4 4000

F = 16000 N

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After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol

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The answer would be to research the need. This should have been done before the project began.

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4 years ago

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An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off

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Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

$V=\frac{\Delta W}{\Delta q}$

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

$I=\frac{\Delta q}{\Delta t}$

Where:

I is the current (2 A)

Δt is the time (20 s)

$2=\frac{\Delta q}{20}$

$q=40 C$

Then, we can put this value of charge in the voltage equation.

$V=\frac{4600}{40}=115 V$

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

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3 years ago

What is the abbreviation for mojave desert?

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3 years ago

A mass weight of 120N is hung from two strings. what is the tension?

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The weight should be shared between the two string equally. Therefore, tension in each string, T is;

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3 years ago

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

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3 years ago