The first step when doing<br> an<br> investigation is the observe a situation. True or false?

miss Akunina [59]

Answer:

yes they can,

Electrostatic forces are non-contact forces; they pull or push on objects without touching them. Rubbing some materials together can result in something called 'charge' being moved from one surface to the other.

Explanation:

example: the opposite ends of magnets

hope this helps! have a wonderfull day!

3 0

3 years ago

WHY is it necessary that the wavelength of this solar radiation (especially in the visible range) be changed so that it can be r

melamori03 [73]

Answer:

radiation reaches the earth it must be absorbed by the different living organisms and when using part of its energy is transformed into radiation of greater wavelength (lower energy) that is irradiated by these organisms; This infrared radiation is absorbed by different gases in the atmosphere

Explanation:

The radiation that comes from the Sun has all wavelengths, due to the 5500K sun temperature its maximum emission is 550 nm (green), when this radiation reaches the earth it must be absorbed by the different living organisms and when using part of its energy is transformed into radiation of greater wavelength (lower energy) that is irradiated by these organisms; This infrared radiation is absorbed by different gases in the atmosphere and cannot escape, creating the so-called greenhouse effect.

If the initial wavelength does not change, it would be emitted into space whereby such temperature of the planet will decrease significantly

8 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0

3 years ago

You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

b)

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$