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3 years ago

What are the order of the 7 prefixes you are required to know in order from largest to smallest

Physics

2 answers:

jekas [21]3 years ago

8 0

Answer:

Giga ( $10^{9}$ ), Mega ( $10^{6}$ ), Kilo ( $10^{3}$ ), centi ( $10^{-2}$ ), milli ( $10^{-3}$ ), micro ( $10^{-6}$ ), nano ( $10^{-9}$ ),

Explanation:

The prefixes to be known is dependent on your level, either a high school, College or at University. Because some other prefixes not listed above are also important to be known, like; Tera ( $10^{12}$ ), femto ( $10^{-15}$ ), pico ( $10^{-12}$ ). But in general, the stated prefixes required to be known with their applications and symbols.

vaieri [72.5K]3 years ago

7 0

10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
Potentially they might want centi which is 10^-2

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A noncontinuous loop that prevents the flow of current is a(n) ____ circuit.

Leona [35]

A noncontinuous loop that prevents the flow of current is a(n) <u>open</u> circuit.

8 0

3 years ago

Read 2 more answers

Assume that charge −q−q-q is placed on the top plate, and +q+q+q is placed on the bottom plate. What is the magnitude of the ele

STatiana [176]

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

q = CV

And

V = q/C ……. (i)

The capacitance is given by the following equation:

C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

V = qd/ Aε0 …..(iii)

The electric field is defined as:

E = V/d …… (iv)

Put equation (iii) in (iv),

E = qd/ Aε0d

E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

8 0

3 years ago

A mixture of helium and argon gas is expanded from a volume of 29.0L to a volume of 82.0L, while the pressure is held constant a

finlep [7]

Answer:

322 kJ

Explanation:

The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.

When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.

For a constant pressure, the work can be calcutated by:

W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:

p = 60 atm = 6.08x10⁶ Pa

ΔV = 82.0 - 29.0 = 53 L = 0.053 m³

W = 6.08x10⁶x0.053

W = 322x10³ J

W = 322 kJ

8 0

3 years ago

¿Por qué cuando agrego cubos de hielo a un vaso de gaseosa, estos se derriten?

77julia77 [94]

Answer:

Esto se debe a que la soda está más caliente que el hielo, lo que hace que pase por su punto de fusión.

Explanation:

Me enteré de esto durante la clase de ciencias hace un par de años. jajaja

4 0

3 years ago

For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago