The rate of a reaction is the speed at which products form or reactants disappear.<br> T/F

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a

Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

$V_c = l^3$

$V_c = 6^3 = 216cm^3$

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

$V_d = \pi r^2*l$

$V_d = \pi (\frac{2}{2})^2*6$

$V_d = 6\pi cm^3$

In this way the net volume would be

$\Delta V = V_c-V_d$

$\Delta V = 216cm^3-6\pi cm^3$

$\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3$

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

$F= mg$

$m = \frac{F}{g}$

$m = \frac{6.6}{9.8}$

$m = 0.673kg$

PART A) From the relation of density as a unit of mass and volume we have to

$\rho = \frac{m}{V}$

$\rho = \frac{0.673}{197.15*10^{-6}}$

$\rho = 3413.64kg/m^3$

PART B) To find the weight of the cube then we apply the ratio of

$W = mg$

$W = V\rho g$

$W = (216*10^{-6})(3413.64)(9.8)$

$W = 7.22N$

3 0

3 years ago

The crankshaft in a race car goes from rest to 3000 rpm in 2.0 s.

a_sh-v [17]

The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions

3 0

2 years ago

Read 2 more answers

If 11,550 J heat is absorbed by a copper object when it is heated from 20.0°C to 30.0°C,

RSB [31]

Answer:

The mass of the copper is 3.00kg

Explanation:

5 0

2 years ago

A truck with a mass of 1.5 x 103 kg accelerates to a speed of 18.0 m/s in 12.0 s from a dead stop. Assume that the force of resi

Yakvenalex [24]

Power = Net Force x velocity
Net force = driving force - force of resistance
Driving force = mass x acceleration
Acceleration = (final velocity - initial velocity) / time
Acceleration = (18 - 0) / 12 = 1.5 m/s²
Driving force = 1.5 x 10³ x 1.5
= 2250 N
Net force = 2250 - 400
= 1850
Power = 1850 x 18
= 3.33 x 10⁴ Watts

3 0

2 years ago

Read 2 more answers

The rate of a reaction is the speed at which products form or reactants disappear. T/F