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3 years ago

Explain what type of substance would resist an electric charge?

Physics

2 answers:

Ksenya-84 [330]3 years ago

8 0

Answer:

rubber/silicone

Explanation: rubber and silicone are found in wire coverings to block the copper wires from contacting other organisms, therefore rubber and silicone is most resistant.

Ksenya-84 [330]3 years ago

3 0

The correct answer is rubber silicone

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What are energy sources

ladessa [460]

Answer:

Solar Energy. The primary source of energy is the sun.

Wind Energy. Wind power is becoming more and more common.

Geothermal Energy.

Hydrogen Energy

Tidal Energy.

Wave Energy.

Hydroelectric Energy.

Biomass Energy.

Explanation:

7 0

3 years ago

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago

Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart n

vfiekz [6]

Answer:

The wavelength is $\lambda = 6.28 *10^{-7}=628 nm$

The frequency is $f = 4.78 Hz$

Explanation:

From the question we are told that

The slit distance is $d = 0.048 \ mm = 4.8 *0^{-5}\ m$

The distance from the screen is $D = 6.50 \ m$

The distance between fringes is $Y = 8.5 \ cm = 0.085 \ m$

Generally the distance between the fringes for a two slit interference is mathematically represented as

$Y = \frac{\lambda * D}{d}$

=> $\lambda = \frac{Y * d }{D}$

substituting values

$\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }$

$\lambda = 6.27 *10^{-7}=628 nm$

Generally the frequency of the light is mathematically represented as

$f = \frac{c}{\lambda }$

where c is the speed of light with values

$c = 3.0 *10^{8} \ m/s$

substituting values

$f = \frac{3.0*10^8}{6.28 *10^{-7}}$

$f = 4.78 Hz$

3 0

3 years ago

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force exerted by the t

vodomira [7]

Complete Question:

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of 1.50 m and a mass of 48.0 kg. If this fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 6.30 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. (a) the salmon's acceleration m/s2 upward (b) the magnitude of the force F during this interval N

Answer:

(a) acceleration = 15.3 ms⁻²

(b) Magnitude of net force = 734.4 N

Magnitude of upward force exerted by tail fin = 1204.8 N

Explanation:

Mass of the salmon fish = 48 kg

Length of the Salmon Fish = 1.5 m

g = 9.8 ms⁻²

(a) salmon's acceleration during the time interval N:

Downward Force on the fish is equal to the Force due to gravity and is given as:

F₂ = mg

= 48 * 9.8

= 470.4 N

The direction of movement of the fish is upward and the acceleration is constant. We are given two different velocities of fish at two different instances.

- When the head breaks out of the water surface first:

Initial velocity = v₁ = 3 m/s

- When two third of its body length is out = d = 1 m

Final Velocity = v₂ = 6.3 m/s

Using the third equation of motion:

2*a*d = v₂² - v₁²

a = (6.3² - 3²)/2*1

a = 15.3 ms⁻²

(b) magnitude of force F during this interval N = ?

We are assuming that F is the net force consisting of both the upward and the downward force.

According to Newton's 2nd law of motion, Force is given as:

F = ma

F = 48 kg * 15.3

F = 734.4 N

Magnitude of upward Force = Fₓ

Force Fₓ exerted by the tail fin of the fish is given by

F = Fₓ - F₂

That is the net force is the sum of the upward and downward forces acting on the fish body. Fₓ is positive because it is in upward direction and F₂ is negative because it is in downward direction. F which is the net force here is positive as Fₓ > F₂.

=> Fₓ = F + F₂

Fₓ = 734.4 + 470.4

Fₓ = 1204.8 N

7 0

4 years ago

According to newton's second law of motion, the force acting on an object varies directly with the object's acceleration. suppos

kolezko [41]

The newton’s second law of motion states that force is directly proportional to acceleration. We can express this mathematically as:

F = k a

where F is the force acting on the object, a is the acceleration of the object and k is a constant of proportionality

Since k is a constant, then we can equate two conditions based on that value:

F1 / a1 = F2 / a2

We are given that:

F1 = 35 N

a 1 = 5 m/s^2

F2 = 49 N

a2 = unknown (?)

Therefore the new acceleration of the object is:

35 / 5 = 49 / a2

a2 = 49 * 5 / 35

a2 = 7 m/s^2

The new acceleration is <span>7 m/s^2.</span>

3 0

4 years ago