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3 years ago

Steroid use in females can cause masculine characteristics, such as lower voice and increased body hair." True False

Physics

1 answer:

s344n2d4d5 [400]3 years ago

7 0

Answer:

True

Explanation:

Side affects can range from

Problems with periods to Loss of breasts

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Mr. Galonski loves to use the Electromagnetic Spectrum. Create a scenario in which Mr. Galonski is using waves from the electrom

Anestetic [448]

Answer:

The Scenario:

On a normal Sunday afternoon Mr. Golanski is sitting in his living room reading his book. He decides after a while to turn on the Television to see what’s on the news, (Mr. Golanski is using Radio waves when he turns his television on by signaling the TV from his remote control). After a few hours Mr. Golanski decides it’s time to have dinner. He heats up a quick meal in his microwave because he doesn’t have the patience for cooking. (He is using microwave radiation to heat his food because water molecules in food absorb the radiation). He sits down for his meal, and halfway through he starts to choke! In a panicked frenzy he runs to his bathroom to try and dislodge the obstacle from his throat. By doing so he switched on the fluorescent lights in his bathroom exposing himself to small amounts of ultraviolet radiation. (Fluorescent lights absorb UV radiation and transmit visible light along with small amounts of UV light). Unable to dislodge the obstacle from his throat Mr. Golanski seeks help from his neighbor who drives him straight to the ER. To treat him properly the physicians opt for a fluoroscopy to examine Mr. Golanski’s esophageal tract. (Thus he is making use of X-ray imaging to obtain a visual of his internal esophageal structure to check for the obstruction). Once treated and discharged from the hospital Mr. Golanski returns home grateful to have survived this ordeal with minimum damage.

Explanation:

The Electromagnetic Spectrum is the range of frequencies and wavelengths for different light waves. They range with increasing frequency from Radio waves, to Microwaves, to Infrared waves, to Visible waves, to Ultraviolet waves, to Infrared waves, to X-rays, and to Gamma rays. Several of which we use in our daily lives such as Radio waves when operating our television or using our cellular phones. We also use microwaves to heat our food or for communication with satellites. We are also exposed to natural Ultraviolet radiation from the sun; however, we can also get exposed to other forms such as from certain types of light bulbs. We see visible light in the form of all the colors we can detect around us. We make use of x-rays for imaging techniques widely used in medicine for diagnostics, as well as Infrared waves in our home security systems. The electromagnetic spectrum is always used as a part of our everyday life.

7 0

3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

3 years ago

A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the

Juli2301 [7.4K]

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

$AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km$

Hence, the van is 4.47 km from its initial point

3 0

2 years ago

A child with mass of 25 kg gets into a toy car with mass of 80 kg on a playground, causing it to sink on its springs (with effec

Bess [88]

Answer:

The frequency of the oscillation is 0.9Hz

Explanation:

This problem bothers on simple harmonic motion of a spring

Given data

Mass of the child m= 25kg

Spring constant k=791 N/m

Amplitude a= 31cm

But the period of the motion as a result of the adults sholve is expressed as

T=2π√m/k

T=2*3.142√25/791

T=6.284√0.031

T=6.284*0.176

T=1.11 sec

But frequency F=1/T

F=1/1.11

F=0.9Hz

4 0

3 years ago