A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and

 Hi!

Instead of providing the answer outright, I'll assist you in grasping it using straightforward concepts.

As electromagnetic waves (consider: light) progress, they oscillate up and down. The frequency of light indicates the number of up-down oscillations occurring each second (1hz means 1 up-down oscillation per second, while 1mhz equals 1 million up-down oscillations per second). Each of these oscillations is known as a single wave.

Since we also have the information about how many meters light covers in 1 second (290,000,000 meters), it's easy to conclude that (regarding this specific question) for every '290,000,000' meters, there are '835,000,000' oscillations or waves.

Calculating the length of a single wave requires a simple division! - which refers to the wavelength of light at the specified frequency.

290,000,000 / 835,000,000 = 0.34m (or 34cm)

Answer:

20.353125 V

Explanation:

m = Mass of proton =

q = Charge of proton =

= Velocity of proton at point A = 50 km/s

= Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

The determined potential difference is 20.353125 V

Answer:

   C = 4,174 10³ V / m^{3/4},  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

In this problem, we are tasked with determining the constant value and the generated electric field.

We will begin with computing the constant C:

           V = C

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

Next, we will find the electric field by utilizing the formula:

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

Substituting x = 0.110 cm:

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × m

dynamic viscosity = 1.75 × Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ

therefore

= µ ............1

substituting the values

= 1.75 × ×

= 0.175 v

the area between the air and puck is given by

Area =

area =

area = 7.85 × m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 ×

force = 1.374 × v

now applying Newton's second law

force = mass × acceleration

- force =

- 1.374 × v =

solving for t =

the time needed after impact for the puck is 2.18 seconds

There's an absence of circuit diagrams.  

Initially, this causes worry for a moment, until we remember that we have no understanding of the experiment mentioned in the problem either, rendering such worries unnecessary.