Ask question

Ask question

All categories

3 years ago

13

Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el

ectric potential. 2. indeterminate; sometimes higher, sometimes lower. 3. lower electric potential. 4. higher electric potential.

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

The electron tends to go to the region of 4. higher electric potential.

Explanation:

When a charged particle is immersed in an electric field, it experiences a force given by

$F=qE$

where

q is the charge of the particle

E is the electric field

The direction of the force depends on the sign of the charge. In particular:

- The force and the electric field have the same direction if the charge is positive

- The force and the electric field have opposite directions if the charge is negative

Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.

However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.

Therefore, the correct answer is

The electron tends to go to the region of 4. higher electric potential.

You might be interested in

Your friend wants to be magician and intends to use Earth’s magnetic field to suspend a current-carrying wire above the stage. H

sesenic [268]

Answer:

$I=1960A$

Explanation:

Under this condition, . In order to suspend the wire, this magnetic force would have to be equal in magnitude to the gravitational force exerted by Earth on the wire the maximum force at angle

F=ILxB

Now to the suspend the wire so use the maximum force and solve to the current knowing the magnetic field of the earth

∑Fy=0

$F_{m}-F_{g}=0$

$I*L*\beta-m*g=0$

Solve to I current

$I=\frac{m*g}{L*\beta}$

$I=\frac{10x10^{-3}kg*9.8m/s^2}{1m*0.5x10{-4}T}$

$I=1960A$

I suggest do an toher act is really risk that current for an act

7 0

3 years ago

The graph represents velocity over time. What is the acceleration? –0.4 m/s2 –0.2 m/s2 0.2 m/s2 0.4 m/s2

Olenka [21]

Missing graph. I attach it in the answer.

In a uniformly accelerated motion, the velocity at time t is given by:
$v(t)=at$
where a is the acceleration and t is the time.

Given the previous equation, if we plot v(t) versus t, we find a straight line; moreover, a (the acceleration) represents the slope of the curve.

Looking at the graph, we see that when the time goes from 10 s to 20 s, the velocity increases from 4 m/s to 6 m/s. Therefore the slope of the curve is
$a= \frac{\Delta v}{\Delta t}= \frac{6 m/s-4 m/s}{20 s-10 s}= \frac{2 m/s}{10 s}=0.2 m/s^2$
and this corresponds to the acceleration.

So, the correct answer is <span>0.2 m/s2.</span>

7 0

3 years ago

Read 2 more answers

Obtain a volume of 12.5 millimeters of liquid in the diagram

tiny-mole [99]

Answer:

D remove 1.5 ML of liquid.

Explanation:

5 0

3 years ago

Read 2 more answers

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

A small cube of metal measures 19.0 mm on a side and weighs 79.6 g. What is the density of the metal in g/cm3?

Nady [450]

Answer:

density of cube =11.605 g/cm³

Explanation:

density of a substance is the mass per unit volume of that substance.

the density of a substance = $\frac{mass}{volume}$

volume of a cube = l³,

l = 19.0mm , lets convert mm to cm

1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm

length of cube =1.9cm

volume of cube = 1.9³

density of cube = $\frac{79.6}{1.9^{3} }$

density of cube =11.605 g/cm³

8 0

3 years ago