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icang [17]
3 years ago
10

A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7

5 m.
Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it?

3.5 m/s
4.1 m/s
13 m/s
16 m/s
Physics
2 answers:
Norma-Jean [14]3 years ago
7 0

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

olga55 [171]3 years ago
4 0

Answer: The correct option is 3.5m/s.

Explanation: In the question, conservation of energy is taking place which means that the energy is getting transferred from one form to another form.

Here, elastic potential energy of the spring is getting converted to the kinetic energy of the block.

Mathematically,

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

where, k = spring constant = 56N/m

x = compression of extension of spring = 0.75m

m = Mass of the block = 2.5kg

v = velocity of the block = ? m/s

Putting values in above equation, we get:

\frac{1}{2}\times 56\times (0.75)^2=\frac{1}{2}\times 2.5\times v^2\\\\v=3.54m/s

Hence, the block will move at a speed of 3.5m/s

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If 400 N of force was applied to a rock and it had an acceleration of 20 ms2 , what is the mass of rock?
Troyanec [42]

Answer:

<h2>20 kg</h2>

Explanation:

The mass of the rock can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{400}{20}  =  \frac{40}{2}  \\

We have the final answer as

<h3>20 kg</h3>

Hope this helps you

5 0
3 years ago
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching e
Degger [83]

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

C=\frac{\epsilon_0.A}{d}

\therefore C\propto \frac{1}{d}

<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>

ii)

As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.

iii)

Electric field between the plates is given as:

E=\frac{\sigma}{\epsilon_0}

where:

charge density, \sigma=\frac{Q}{A}

<em>As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.</em>

iv)

  • From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.
  • Here we increase the distance so the work done per unit charge increases.
6 0
3 years ago
A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t
Marrrta [24]

Answer:

a) U = 652.545\,kJ, b) v \approx 659.568\,\frac{m}{s}

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

Q = U

The internal energy for a monoatomic ideal gas is:

U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

T = \frac{P\cdot V}{n\cdot R_{u}}

T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}

T = 52.325\,K

The thermal energy contained by the gas is:

U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)

U = 652.545\,kJ

b) The physical model for the cat is constructed from Work-Energy Theorem:

U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

v = \sqrt{\frac{2 \cdot U}{m_{cat}}}

v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }

v \approx 659.568\,\frac{m}{s}

3 0
3 years ago
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dimaraw [331]

Answer:

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san4es73 [151]

Answer:

Information

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Explanation:

6 0
3 years ago
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