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4 years ago

2. In regards to the the zonal anatomy of the prostate gland, where do most carcinomas originate?

Physics

1 answer:

AveGali [126]4 years ago

7 0

Answer:

The peripheral zone is in the outer most part of the prostate, and the lower peripheral zone is fairly close to the rectal wall. The peripheral zone is the most common site for prostatic adenocarcinoma. The central zone is in the center of the prostate and cancer does not originate there often.

Explanation:

The peripheral zone (PZ) contains the majority of prostatic glandular tissue. ...

The central zone (CZ) is the area that surrounds the ejaculatory ducts. ...

The transition zone (TZ) surrounds the urethra as it enters the prostate gland.

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What is the speed of a 200-kilogram car that is driving with 2000 joules of kinetic energy? (SHOW ALL WORK)

Katyanochek1 [597]

Answer:

v ≈ 4.47

Explanation:

The Formula needed = KE = $\frac{1}{2}$ m v²

Substitute with numbers known:

2000J = $\frac{1}{2}$ × 200kg × v²

Simplify:

÷100 ÷100 (Divide by 100 on both sides)

2000J = 100 × v²

$\frac{2000J}{100}$ = v²

20 = v²

√ √ (Square root on both sides)

√20 = √v²

4.472135955 = v (Round to whatever the question asks)

v ≈ 4.47 (I rounded to 2 decimal places or 3 significant figures, as that is what it usually is)

3 0

3 years ago

This is a computer program, why is there an error between the computer values and the nominal values of R?

ss7ja [257]

Answer:

the difference is due to resistance tolerance

Explanation:

In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.

This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values and the values measured with the instruments, multimeters.

Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost

In summary, the difference is due to resistance tolerance.

5 0

4 years ago

lord [1]

An inclined plane makes work easier by breaking an upward or downward movement into smaller increments.

A screw is like an inclined plane wrapped around a cylinder. It turns a small rotational force into a larger forward driving force

7 0

3 years ago

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$ .

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$ , then $\frac{v_{2}}{v_{1}} > 1$ . That is to say, $v_{1} < v_{2}$ .

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$ , then $p_{1} > p_{2}$ . Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.

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3 years ago

A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are app

lys-0071 [83]

The car's velocity at time t is given by

$v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t$

It comes to a stop when v = 0, which happens when

$0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s$

or after about 13.9 s.

In this time, the car travels a distance x given by

$x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m$

or about 192 m.

In one complete revolution, each tire covers a distance equal to its circumference,

$2\pi(0.340\,\mathrm m)\approx2.13628\,\mathrm m$

or about 2.14 m.

This means each tire will complete approximately 192/2.14 ≈ 90 revolutions.

8 0

3 years ago