As a result of photosynthesis is the production of sugar molecules known as what?

In practice, both the rivet itself (length, diameter) and the hole in which it is placed have tolerances. How do these affect th

Nikolay [14]

Answer:

The quality of a formed rivet is determined by the geometry from its formation of the river heat, and the extent to which the hole is filled up by the rivet.

Explanation:

The quality of a formed rivet is determined by the geometry from its formation of the river heat, and the extent to which the hole is filled up by the rivet. This helps determine the squeeze force, rivet length, rivet diameter tolerance, hole countersunk depth and hole diameter tolerance, plays a big role on the quality of rivet formed. When the tolerance of the hole it’s placed is smaller the grip from the river is stronger unlike when it’s loose.

8 0

3 years ago

Zoning laws establish _______.

Aleonysh [2.5K]

Answer:

Zoning ordinances detail whether specific geographic zones are acceptable for residential or commercial purposes. Zoning ordinances may also regulate: - size

- placement

- density

- height of structures

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

5 0

3 years ago

List out any 4 objectives of education

e-lub [12.9K]

- vocational aim
- cultural aim
- spiritual aim
- intellectual aim

6 0

2 years ago

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

4 years ago

Read 2 more answers

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\\h=19.62 meters$

Part b

$v=u+at\\v=0+9.81*2\\v=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs$

4 0

3 years ago