What is the name of the process that describes the decrease in frequency

What is the potential difference (voltage) if the resistance of 20 ohms produces a current of 10 amperes?

JulijaS [17]

Answer:

200 volts

Explanation:

Ohm's Law: V = IR

I = 10 A

R = 20 ohms

V = (10)(20) = 200 volts

8 0

2 years ago

In which of the following situations would it NOT be wise to estimate?

ipn [44]

Medicine to a patient. That should be calculated based on weight, strength/dosage and possibly other factors

5 0

2 years ago

Which statement about stage 3 NREM is accurate?

cestrela7 [59]

Answer:

hope

that helps

Explanation:

D. The sleeper's heart rate, blood pressure, and breathing rate drop to their lowest levels.

8 0

2 years ago

Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s

Vitek1552 [10]

The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for 52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=binding energy
52He-----------> 2 x 11p + 3 x10n is the equation bilan
so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=1.67 10^-27 kg
 m nucleus= 5
Dm= 2x1.67 10^-27 kg+ 3x1.67 10^-27 kg-5= - 4.9 J
Eb= - 4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so -9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV

the final answer is Eb /nucleon = -5.625 x10^35 eV

8 0

2 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

2 years ago