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2 years ago

What is the magnitude of the electric field at the dot in the figure? (figure 1)

Physics

1 answer:

blondinia [14]2 years ago

5 0

The magnitude of the electric field at the dot is : 10⁴ v/m

<h3>What is electric field?</h3>

Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

Determine the magnitude of the electric field at the dot

change in voltage = E .d

$100 - 0 = E \times ( 1 \times 10^{-2}\ m ) ----- ( 1 )$

From equation ( 1 )

The magnitude of electric field will be given as

$E = \dfrac{100 }{ ( 1 \times 10^{-2})\ }$

$E = 10^4\ \ \frac {v}{ m}$

Hence we can conclude that The magnitude of the electric field at the dot is : 10⁴ v/m

Learn more about electric field :

brainly.com/question/14372859

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A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring project

Gwar [14]

Answer:

$x' = 10 x$

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

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

so we will have

$v = \sqrt{\frac{k}{m}}(x)$

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

$5v = \sqrt{\frac{k}{4m}}(x')$

now from above two equations we have

$\frac{5v}{v} = \frac{x'}{2x}$

so we have

$x' = 10 x$

3 0

2 years ago

A 4.00kg counterweight is attached to a light cord, which is would around a spool. The spool is a uniform solid cylinder of radi

Sergio [31]

Answer:

Explanation:

Given that,

Mass of counterweight m= 4kg

Radius of spool cylinder

R = 8cm = 0.08m

Mass of spool

M = 2kg

The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:

Then we have,

τ(net) = R~ × T~

τ(net) = R~•i × mg•j

τ(net) = Rmg• k

τ(net) = 0.08 ×4 × 9.81

τ(net) = 3.139 Nm •k

The magnitude of the net torque is 3.139Nm

b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:

L~ = R~ × m~v + I~ω

L = mRv + MR v

L = (m + M)Rv

L = (4 + 2) × 0.08

L = 0.48 Kg.m

C. τ =dL/dt

mgR = (M + m)R dv/ dt

mgR = (M + m)R • a

a =mg/(m + M)

a =(4 × 9.81)/(4+2)

a = 6.54 m/s

6 0

3 years ago

Read 2 more answers

If an electric charge is accumulated on an object, it is referred to as _____

LenaWriter [7]

Static, because it stays on the object.

3 0

3 years ago

Read 2 more answers

A cook puts 1.90 g of water in a 2.00 L pressure cooker that is then warmed from the kitchen temperature of 20°C to 111°C. What

sergey [27]

Answer:

P= 168258.30696 Pa

Explanation:

Given that

Mass of water vapor m = 19.00 g

Volume of water vapor V = 2.00 L

$=2.00\times10^{-3}m^3$

Temperature of water vapor is T = 111°C

= 384K

Molar mass of water is M = 18.0148 g/mol

Number of moles are

n = m/M

= (1.90 g)/(18.0148 g/mol)

= 0.1054 mol

Pressure inside the container is

P= nRT/V

$P=\frac{(0.1054)(8.314472)(384)}{(2.00\times10^{-3})}$

P= 168258.30696 Pa

3 0

3 years ago

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient

morpeh [17]

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ \frac{t}{9}}$

Explanation:

When no sliding friction and no air resistance occurs:

$m\frac{dv}{dt} = mgsin \theta$

where;

$\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}$

Taking m = 3 ; the differential equation is:

$3 \frac{dv}{dt}= 128*\frac{1}{2}$

$3 \frac{dv}{dt}= 64$

$\frac{dv}{dt}= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}$

$\frac{dv}{dt}= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v$

= $3 \frac{dv}{dt}=16 -\frac{1}{3}v$

By integration

$v(t) = 48 + Ce ^{\frac{t}{9}$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ \frac{t}{9}}$

7 0

2 years ago