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Wittaler [7]
3 years ago
9

An electrical field has: direction but not magnitude magnitude but not direction neither magnitude nor direction none of the abo

ve
Physics
2 answers:
Elanso [62]3 years ago
3 0
An electric field has both magnitude and direction.
I don't think that's one of the first 3 choices.
Dovator [93]3 years ago
3 0

Answer:

Explanation:

The region around a charge particle in which another charged particle experiences a force of attraction or repulsion is called electric field.

The strength of electric field is defined as the force experienced by the unit charge.

E = F / q

As force is a vector quantity, it has direction and magnitude both, so electric field is also a vector quantity.

Thus, electric field has magnitude and direction both.

You might be interested in
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
A 7.2 magnitude earthquake, with an epicenter in Northern Mexico, was felt over 100 miles away in Southern California because wa
ryzh [129]

Answer:

B

because waves generated

7 0
1 year ago
1. In this activity, you will be looking for a relationship between the mass of the cart and the acceleration of the cart.
posledela

Answer:

Mass of cart

Explanation:

7 0
3 years ago
A spacecraft left Earth to collect soil samples from Mars. Which statement is true about the strength of Earth’s gravity on the
ryzh [129]

Answer:

: It Decreases.

As the spacecraft gets farther and farther from Earth, the gravitational  

forces between the spacecraft and the Earth decrease.

Explanation:

5 0
3 years ago
At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa
ivann1987 [24]

Answer:

(b) EAST

Explanation:

you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:

\vec{B}=B_o\hat{i}\\\\\vec{v}=v_o\hat{j}

You use the Lorentz formula to known which is the direction of the magnetic force over the electron:

F=qv\ X\ B

which implies the cross product between the unitary vecors j and i, that is

\hat{i} \ X\ \hat{j} = -\hat{k}  (WEST)

However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST

3 0
3 years ago
Read 2 more answers
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