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3 years ago

8

_______________ appears on the ecg as having no p wave, a wide qrs complex, and t waves that deflect in the opposite direction f

rom the r wave.

1 answer:

melomori [17]3 years ago

5 0

It would probably be <span>Premature Ventricular Contractions.

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In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ

grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E $_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E $_{max$ )

E $_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E $_{max$ = 350/60 kV/m

E $_{max$ = 5.83333 kV/m

E $_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E $_{max$ = 5833.33 N/C

so we substitute all our values into the formula for intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0

3 years ago

Two passenger cars, car A and car B, of the same weight are put to a 30 mph head-on crash test. In the test, cars are pulled for

maksim [4K]

Answer:

the statements the correct one is A

Explanation:

Let's analyze this exercise, vehicles have the same mass and speed, so we can use the momentum impulse ratio

I = ∫ F dt = Δp

the Δp is the same for both cars since they have the same mass and the same speeds, so the momentum is the same in both vehicles

When they indicate that vehicle A was reduced more than vehicle B, this implies that the force acted for a longer time, to have the largest reduction in size, therefore the impact force was less in car A than in car B

Resisting the statements the correct one is A

7 0

3 years ago

प्रकाशको आवर्तनको परिभाषा लेख्नुहोस् । कन्भेक्स ऐनाको अगाडि तपाई उभियौ भने तपाईको कस्तो आकृति बन्छ, लेख्नर

allochka39001 [22]

A diagram alright of the

4 0

3 years ago

Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative

Nastasia [14]

Answer:

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Explanation:

It is given that,

$\vec{u}=2i-j-3k$

$\vec{v}=-3i+j-2k$

Taking the cross product of v and v such that,

$\vec{w}=u\times v$

$\vec{w}=(2i-j-3k)\times (-3i+j-2k)$

$\vec{w}=5i+13j-k$

$|w|=\sqrt{5^2+13^2(-1)^2}$

|w| = 13.92

Let $\hat{w}$ is the unit vector normal to the plane containing u and v. So,

$\hat{w}=\dfrac{\vec{w}}{|w|}$

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Hence, this is the required solution.

7 0

4 years ago

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

4 years ago