8 days ago

The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at

temperature of 20 0 c and 50 c, respectively. determine the rate of heat transfer through the wall, in w.

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

ValentinkaMS [3465]

Answer:

The particle's acceleration magnitude and direction is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given:

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

To find the particle’s acceleration

The formula for acceleration is

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We will calculate $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, substitute all known values into the formula for acceleration

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

The negative sign indicates the direction is opposite.

Thus, the particle’s acceleration magnitude and direction is $a= 0.3296\ \hat{k}\ m/s^2$

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(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

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A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?

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Answer: 24.24 m

Explanation:

A player launches a football 50.0 m at an angle of 61° to the north of west. We will break this down into vertical and horizontal elements.

Horizontal component: 50 cos 61° = 24.24 m directed westward

Vertical component: 50 sin 61° = 43.73 m directed toward the north.

Refer to the diagram below.

Therefore, the westward displacement of the football corresponds to the horizontal component of the displacement, which is 24.24 m.

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