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Debora [2.8K]
3 years ago
10

7.In North America, a magnetic compass will point to

Physics
1 answer:
Mashcka [7]3 years ago
6 0

Answer:

I think then answer for this question is number c. the South pole

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If a molecule has four hybrid sp3 orbitals, it can be concluded that the molecule has a
Lilit [14]
If a molecule is found to have four hybrid sp3 then we say that it has a <span>tetrahedral molecular geometry. thsi is the correct term to define this type of molecules. So your option would be C. Hope this is useful</span>
7 0
3 years ago
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A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.
Alexxandr [17]

Answer:

Explanation:

Given that,

B(t) = B0 cos(ωt) • k

Radius r = a

Inner radius r' = a/2 and resistance R.

Current in the loop as a function of time I(t) =?

Magnetic flux is given as

Φ = BA

And the Area is given as

A = πr², where r = a/2

A = πa²/4

Then,

Φ = ¼ Bπa²

Φ(t) = ¼πa²Bo•Cos(ωt)

Then, the EMF is given as

ε(t) = -dΦ/dt

ε(t) = -¼πa²Bo • -ωSin(ωt)

ε(t) = ¼ωπa²Bo•Sin(ωt)

From ohms law,

ε = iR

Then, i = ε/R

I(t) = ¼ωπa²Bo•Sin(ωt) /R

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Check attachment for better understanding

7 0
3 years ago
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if
KengaRu [80]

Answer:

The value of  the correct angle of banking for the road is \theta = 67.76 °

Explanation:

Given data

Velocity (v) = 60 \frac{m}{s}

Radius = 150 m

The velocity of the car in this case is given by

v = \sqrt{r g \tan \theta}

v^{2} = r g \tan \theta

\tan \theta = \frac{v^{2} }{rg}

Put all the values in above formula we get

\tan \theta = \frac{60^{2} }{(150)(9.81)}

\tan \theta = 2.446

\theta = 67.76 °

Therefore the value of  the correct angle of banking for the road is \theta = 67.76 °

4 0
3 years ago
Need help 8th grade science test need help
jolli1 [7]
ANSWER: d, average speed
7 0
2 years ago
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