6+8t+3t²=0<br> find time

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

7 0

3 years ago

A scientific law is based on repeated inferences made over long periods of time. Please select the best answer from the choices

sergiy2304 [10]

It would be true
I believe

5 0

3 years ago

Read 2 more answers

A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

zavuch27 [327]

Answer:

Initial velocity will be 1.356 m/sec

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity $u_x=ucos\Theta =ucos37^{\circ}=0.7986u$

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time $t=\frac{distance}{speed}$

$1.20=\frac{1.3}{0.7986u}$

$u=1.356m/sec$

So initial velocity will be 1.356 m/sec

3 0

2 years ago

We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat

Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

$Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)$

Here, mw is the mass of water

$Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal$

Now, you need to calculate the heat released by the steam:

$Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal$

The work done by this engine is the difference between both heats:

$W=Q_{1}-Q_{2}=62760-61960=800cal$

8 0

3 years ago

6+8t+3t²=0 find time​

6+8t+3t²=0 find time