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3 years ago

Which of the following quantities are unknown?

Physics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:

unknow e and f

Explanation:

In experiments with alpha particles that are obtained by the method of radioactive decay of atoms, some parameters are known

a) Known. The initial velocity is given by the energy of the particles entities by the atomic nuclei

b) Known. The particle charge always 2e, helium core

c) Known. It is set in the given experiment, in general it is selected as zero

d) Known. Placed by the experimenter

e) Unknown. The speed depends on the interactions with the system

f) Unknown. It depends on the interactions with the system, because the position depends on the interactions

g) Known. It is always the value of a helium nucleo

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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is

kodGreya [7K]

3.5m is ur answer ask for more questions anytime

5 0

3 years ago

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option

3 0

3 years ago

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

$\Delta E = -\frac{dV}{dr}$

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

$\Delta E = 0$

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0

3 years ago

I need help with this problem :(

Reil [10]

Answer:

it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.

7 0

2 years ago