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Nana76 [90]
3 years ago
7

What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C

Physics
2 answers:
hammer [34]3 years ago
6 0

Answer:

3.58*10⁻¹³ C

Explanation:

When we have a net charge on a sphere, this means that there must be a difference in the number of electrons and protons on the sphere, otherwise, the sphere would be electrically neutral.

In this case, we can find this difference as follows:

Np -Ne = 8.11*10⁶ protons - 5.87*10⁶ electrons = 2.24*10⁶ protons.

The total charge carried by all these protons is just the product of the charge of only one proton (which is equal to the elementary charge e) times the number of excess protons, as follows:

Qnet = 2.24*10⁶ protons * 1.6*10⁻19 C/proton = 3.58*10⁻¹³ C

AysviL [449]3 years ago
3 0

Answer:

3.6 * 10^-13 C

Explanation:

The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,

Q = Qe + Qp

TOTAL ELECTRON CHARGE:

An electron has an electronic charge of -1.6022 * 10^-19C.

Hence, the charge of 5.87 * 10^6 electrons will be:

Qe = - 1.6022 * 10^-19 * 5.87 * 10^6

Qe = - 9.405 * 10^-13 C

TOTAL PROTON CHARGE:

A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:

Qp = 1.6022 * 10^-19 * 8.11 * 10^6

Qp = 1.3 * 10^-12 C

=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)

Q = 3.6 * 10^-13 C

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Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f
musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

E_1_x = 0V/m\\E_1_y=100V/m

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m

∑x: E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m

∑y: E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees

or 27.23° to the left of E1.

8 0
2 years ago
Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

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