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3 years ago

The electromagnetic wave that CT scans are based on is called a

Physics

1 answer:

kherson [118]3 years ago

6 0

X-ray)
because Electromagnetic waves are in act

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Two men are pushing a box in the same direction, to the right along the floor. the first man is pushing with a force of 20n. the

SIZIF [17.4K]

20N + 10N - 5N = 25N. That's the net force.

6 0

3 years ago

Completenlo por favor

Ainat [17]

La velocidad $\mathbf v$ del objeto al tiempo $t$ es descrito por

$\mathbf v(t)=v_x(t)\,\mathbf i+v_y(t)\,\mathbf j$

donde

$\begin{cases}v_x(t)={v_x}_0+a_xt\\v_y(t)={v_y}_0+a_yt\end{cases}$

El objeto no tiene aceleración horizontal, pues $a_x=0$ y $v_x$ está determinado exactamente por su velocidad inicial en la dirección horizontal. En la dirección vertical, la gravedad es la única fuerza que actúa en el objeto, pues $a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}$ . Entonces, la velocidad después de $3\,\mathrm s$ satisface

${v_x}_0\,\mathbf i+\left({v_y}_0+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3\,\mathrm s)\right)\,\mathbf j=20\,\mathbf i-4\,\mathbf j$

Inmediatamente, vemos que ${v_x}_0=20\,\dfrac{\mathrm m}{\mathrm s}$ y podemos encontrar que ${v_y}_0=25.43\,\dfrac{\mathrm m}{\mathrm s}$ .

Su posicíon es descrita al tiempo $t$ por

$\mathbr r(t)=r_x\,\mathbf i+r_y\,\mathbf j$

con

$\begin{cases}r_x={r_0}_x+{v_0}_xt+\dfrac12a_xt^2\\\\r_y={r_0}_y+{v_0}_yt+\dfrac12a_yt^2\end{cases}$

Si la posición inicial del objeto es el origen, suponemos que $\mathbr r(0)=\mathbf 0$ , y además tenemos

$\mathbf r(t)=\left(20\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\mathbf i+\left(\left(25.43\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right)\,\mathbf j$

Queremos determinar el máximo de $r_y$ . Encontramos que ${r_y}_{\mathrm{max}}\approx32.9\,\mathrm m$ cuando $t\approx2.59\,\mathrm s$ .

3 0

3 years ago

Using a rope that will snap if the tension in it exceeds 379 N, you need to lower a bundle of old roofing material weighing 473

Tema [17]

Answer:

Acceleration = a = -1.94 m/s^2

Explanation:

Weight of the bundle = W = mg

m = W/g = 473/9.8

m = 48.26 kg

There are two forces acting on the bundle such as tension in the upward direction and weight in the downward direction. So, according to newton’s second law of motion:

Sum of forces = ma

T - W = ma

a = (T – W)/m

a = (379 - 473)/48.26

a = -1.94 m/s^2

Negative sign shows that the acceleration is in the downward direction.

8 0

3 years ago

A calf stretch is a/an _______________________________.

swat32

<span>C anaerobic activity </span>

5 0

3 years ago

The rate an object is moving relative to a reference point is its

natulia [17]

Velocity is an object's rate of change (usually over time) of its displacement.

5 0

3 years ago