Napoleon's army captured the island of San Salvador, then moved on to occupy Louisiana and close the port of New Orleans.

If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?

Nataly_w [17]

In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C

8 0

3 years ago

A stalled car is being pushed up a hill at constant velocity by three people. The net force on the car is Group of answer choice

IgorLugansk [536]

Answer:

Not understanding question

Explanation:

Mark me brainlist

5 0

2 years ago

A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

$Torque = \vec{r} \times \vec{F}$

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

$Torque = \vec{r} \times \vec{F}$

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :