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tekilochka [14]
3 years ago
9

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

ant value of 316N/m, until the spring is compressed by 35cm. The block is then released. What is the block's final speed, in m/s?
Physics
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

U=\frac{1}{2}kx^2

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

K=\frac{1}{2}mv^2

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s

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The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
sammy [17]

Answer:

(a) emf_L=-LI_{max}\omega cos(\omega t)

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

I(t)=I_{max}sin(\omega t)   (1)

(a) The induced emf is given by the following formula

emf_L=-L\frac{dI}{dt}    (2)

You derivative the expression (1) in the expression (2):

emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

emf_L=-\omega LI_{max}

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

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2 years ago
A rigid, insulated tank contains steam at 3 MPa, 400 Celsius. A valve on the tank is opened, allowing steam to escape. The overa
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Answer:

x=0.8

Explanation:

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p_1=3Mpa\\T_1=400\° c\\q_{1-2}=0

Where p_1 is the pressure of the super heated steam

T_1 is the Temperature of the super heated steam

and q_{1-2} is the pressure in an adiabatic process.

State 1

v_1 = 0.09936m^3/Kg\\s_1=6.9213kJ/kg.K

State 2

s_2=s_1=6.9212

Note that here in state 2 the process is Reversible.

At the value where T_2 = 141\°c we have

v_2=(v_g)_{T_2}=0.4972m^3/kg

Through this values we can calculate the Fraction of steam,

x=\frac{m_1-m_2}{m_1}\\x=1-\frac{m_2}{m_1}\\x=1-\frac{v_1}{v_2}\\x=1-\frac{0.9936}{0.4972}\\x=0.8

4 0
3 years ago
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