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3 years ago

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A point charge Q moves on the x-axis in the positive direction with a speed of 370 m/s. A point P is on the y-axis at y = +80 mm

. The magnetic field produced at point P, as the charge moves through the origin, is equal to 0.80x10^6 T in the NEGATIVE z direction . When the charge is at x = +40 mm, what is the magnitude of the magnetic field at point P?

1 answer:

mixas84 [53]3 years ago

3 0

Answer:

+58

Explanation:

Cause I intelligent

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A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87 = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

$n=\dfrac{m}{M}$

M=Molar mass

$n=\dfrac{m}{M}$

$0.99=\dfrac{28.6}{M}$

M=28.88 gm/mol

3 0

3 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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3 years ago

si tienes 2 circuitos uno en serie y otro en paralelo con el mismo valor en sus resistencias, para cuál circuito la resistencia

Novosadov [1.4K]

Answer:

circuito paralelo

Explanation:

Siempre el circuito en paralelo dara una resistencia menor. Recuerda que las resistencias se suman en el circuito en serie, an cambio en el circuito en paralelo, la corriente se bifurca de manera de circular con mayor intensidad por las ramas que tengan menos resistencia, y tal situacion llevara siempre a producir una menor resistencia equivalente.

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When heat flows from one substance to another, what happens to the temperature of the substance giving off the heat and to the t

blagie [28]

The temperature of the substance giving off the heat decreases while the temperature of the substance receilving the heat increases. they leach what is called equlibrium point where heat energy can longer be exchanged hence equql temperature. this isThermal physics

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Which material rises from cracks in oceanic crust?

disa [49]

Answer:

trenchs and magma

Explanation:

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