One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

Answer:

a)n= 3.125 x electrones.

b)J= 1.515 x A/m²

c) =1.114 x m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x m

radio 'r' = d/2 => 1.025 x m

número de electrones 'n'= 8.5 x

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x C

n= Q/e => 5/ 1.6 x

n= 3.125 x electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x )²)

J= 1.515 x A/m²

c) La velocidad típica '' de un electrón se expresa como:

=

    =1.515 x / 8.5 x x |-1.6 x |

=1.114 x m/s

d) De acuerdo con estas ecuaciones,

J= I/A

= =

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>

The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.

Answer:

(a). The coefficient of permeability is .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Insert the values into the formula

(a). We will compute the coefficient of permeability

Applying the formula for permeability

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

The coefficient of permeability measures as .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

Substituting the values into the equation

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

Answer:T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42 ksi = 42 x 6.89 = 289.58 MPa

To find Torque = T =?

Solution:σ = (P x r) / T

Thus, T = (P x r) / σ

Calculating T = (386.11 x 0.02) / 289.58

results in T = 0.03 Nm.