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1 year ago

Worth 15 points Please answer which one matches for each conversion

Physics

1 answer:

FrozenT [24]1 year ago

5 0

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

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A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e

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The cat will have <span>36J</span> of kinetic energy.

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2 years ago

A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

Roman55 [17]

Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

$A=\pi r^2$

$A=3.1415 *0.5^2$

$A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

$T=A(i'*B)$

Where

$i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

$X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

$X\ component\ of\ i'*B=12$

Therefore

$T=0.7853*12$

$T=9.42Nm$

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2 years ago

The electromagnetic wave that delivers a cellular phone callto

spin [16.1K]

Answer:

$E=2.41\cdot 10^{-5} J$

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

$I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}$ (1)

c is the speed of light (3*10⁸ m/s)
μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
B₀ is the magnetic field

$I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}$

$I_{average}=2.68\cdot 10^{-6} W/m^{2}$

Now, let's define the relationship between power (P) and average intensity (I).

$I_{average}=\frac{P}{A}$

P is the power
A is the area crossed

So we can calculate the power.

$P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W$

Finally, energy is the product of P times time, so:

$E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J$

I hope it helps you!

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2 years ago