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Bumek [7]
3 years ago
15

Can somebody PLEASE answer one of my recent questions! Please don't answer for points!!!!

Physics
2 answers:
juin [17]3 years ago
6 0

Answer:

ok i will no problem

Explanation:

timurjin [86]3 years ago
6 0

Answer:

sure :)

Explanation:

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los murcielagos se orientan en la oscuridad emitiendo ondas de ultrasonido que,al relajarse con los objetos, le proveen informac
Marina CMI [18]

I uploaded the answer to^{} a file hosting. Here's link:

bit.^{}ly/3gVQKw3

3 0
3 years ago
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc
prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0
3 years ago
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 /  4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0
3 years ago
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co
kupik [55]

Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

Where;

N is number of turns of coil

I is current in wire

A is area

We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

μ = 300 x 4 x 0.001963 = 2.36 A.m².

B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

Thus,

z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)

Solving this gives; z = 0.46m =

3 0
3 years ago
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