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3 years ago

Anyone know how to solve this? 4+9×2÷3−1

Physics

1 answer:

Nutka1998 [239]3 years ago

5 0

You want to use PEMDAS to solve this equation.

4+9*2/3-1

4+18/6-1

4+6-1

10-1

Your answer is 9

Thanks -John

If you have anymore question just ask! :)

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Which type of central heating system involves fans and ducts to circulate warm air?

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Forced-Air would be the type of central heating system

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3 years ago

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage s

miskamm [114]

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

σ = n e2 τ / m*

The relationship with resistivity is

ρ = 1 /σ

Whereby the resistance

R = ρ L / A = 1 /σ L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

T = 2 T₀

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4 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago

An airplane is heading south at a speed of 600km/h. If a wind begins blowing from the southwest at a speed of 100km/h, calculate

Karo-lina-s [1.5K]

Answer:

Here's what I got:

Let's assume that N and E are + directions while S and W are - directions.

Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).

Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h

Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h

Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71

Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h

To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg

ANSWER: velocity = 533.99 km/h at 82.39 deg S of E

Explanation:

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4 years ago