PLEASE HELP WILL GIVE BRAINLIEST

If you pluck a guitar string hard the sound waves it gives off will have __________ amplitudes.

Ilia_Sergeevich [38]

Answer:

large

Explanation:

the harder you pluck the louder it gets

4 0

3 years ago

Read 2 more answers

If electrical current is moving through a horizontal wire toward your face, what direction is the induced magnetic field?

adoni [48]

Answer:(b)

Explanation:

Given

Electric current flowing through the horizontal wire towards the observer's face.

The direction of the magnetic field is given by the right-hand thumb rule, i.e. place the thumb in the direction of current and the wrapping of fingers will give the direction of the magnetic field

the direction of the magnetic field will be counterclockwise as observed by an observer.

4 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago

Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

8 0

3 years ago

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x

Ostrovityanka [42]

Answer:

(a) increase

Explanation:

On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.

The change in the potential energy of a charge field-system can be given as:

$\delta U= -q(EdsCos \theta)$

where;

q = positive test charge

E = Electric field

ds = displacement between thee charge positions

θ = Angle between the electric field and the displacement.

Given that:

Charge of the particle = -q

displacement = (60.0 -20.0)cm = 40.0 cm

θ = 0

Replacing our values in the above equation, we have:

$\delta U = -(-q)(60Cos 0)$

$\delta U = qEds$

Since the potential energy of the system is positive, therefor the electric potential energy also increases.

5 0

3 years ago