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3 years ago

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1. Calculate the kinetic energy of a moving object, its mass= 100 kg, and its velocity= 12 m/s.

1 answer:

alexira [117]3 years ago

3 0

k=1/2mv2

k=1/2 100kg*12m/s2

k=7200

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An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of

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Explanation:

$2\sqrt{34}$

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What is a linear motion

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Linear motion (also called rectilinear motion) is a motion along a straight line, and can therefore be described mathematically using only one spatial dimension.

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What does Newton's third law describe?

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Answer:

Newton's third law is: For every action, there is an equal and opposite reaction.

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3 years ago

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Three forces act on a statue. Force F⃗ 1F→1 (magnitude 45.0 NN) points in the +x-direction, Force F⃗ 2F→2 (magnitude 105 NN) poi

Trava [24]

Answer:

Resultant force = (232.93î + 246.10j) N

x-component of the resultant force = (+232.93î) N

y-component of the resultant force = (+246.1j) N

Explanation:

The net external force on the statue is equal to the resultant force on the statue.

And the resuphant force is a vector sum of all the other forces acting on the statue.

Force 1 = (45î) N

Force 2 = (105j) N

Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N

Resultant force = (Force 1) + (Force 2) + (Force 3)

Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N

Hope this helps!!!

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3 years ago

Linear Thermal Expansion (in one dimension)

Leokris [45]

Answer:

1) $\Delta L= 0.612\ m$

2) a. $\Delta V_G=0.57\ L$

b. $\Delta V_S=0.021\ L$

c. $V_0=0.549\ L$

Explanation:

1)

given initial length, $L=1275\ m$

initial temperature, $T_i=-15^{\circ}C$

final temperature, $T_f=25^{\circ}C$

coefficient of linear expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

<u>∴Change in temperature:</u>

$\Delta T=T_f-T_i$

$\Delta T=25-(-15)$

$\Delta T=40^{\circ}C$

We have the equation for change in length as:

$\Delta L= L.\alpha. \Delta T$

$\Delta L= 1275\times 12\times 10^{-6}\times 40$

$\Delta L= 0.612\ m$

2)

Given relation:

$\Delta V=V.\beta.\Delta T$

where:

$\Delta V$ = change in volume

V= initial volume

$\Delta T$ =change in temperature

initial volume of tank, $V_{Si}=60\ L$

initial volume of gasoline, $V_{Gi}=60\ L$

initial temperature of steel tank, $T_{Si}=15^{\circ}C$

initial temperature of gasoline, $T_{Gi}=15^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_G=950\times 10^{-6}\ ^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_S=35\times 10^{-6}\ ^{\circ}C$

a)

final temperature of gasoline, $T_{Gf}=25^{\circ}C$

∴Change in temperature of gasoline,

$\Delta T_G=T_{Gf}-T_{Gi}$

$\Delta T_G=25-15$

$\Delta T_G=10^{\circ}C$

Now,

$\Delta V_G= V_G.\beta_G.\Delta T_G$

$\Delta V_G=60\times 950\times 10^{-6}\times 10$

$\Delta V_G=0.57\ L$

b)

final temperature of tank, $T_{Sf}=25^{\circ}C$

∴Change in temperature of tank,

$\Delta T_S=T_{Sf}-T_{Si}$

$\Delta T_S=25-15$

$\Delta T_S=10^{\circ}C$

Now,

$\Delta V_S= V_S.\beta_S.\Delta T_S$

$\Delta V_S=60\times 35\times 10^{-6}\times 10$

$\Delta V_S=0.021\ L$

c)

Quantity of gasoline spilled after the given temperature change:

$V_0=\Delta V_G-\Delta V_S$

$V_0=0.57-0.021$

$V_0=0.549\ L$

8 0

3 years ago