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3 years ago

9

Go to the roller coaster simulation and click on the “launch” button. Pay attention to the pie chart as the roller coaster moves

and fill in the blank. Be sure to keep this website open because there will be several questions related to it. The kinetic energy as the roller coaster goes downhill.

2 answers:

brilliants [131]3 years ago

4 0

Answer:

increases

Explanation:

tekilochka [14]3 years ago

3 0

Answer:

increases

Explanation:

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Why would machine-operated stopwatches be used at sports events instead of hand-operated stopwatches?

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The grains found in igneous rock:

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

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3 years ago