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3 years ago

A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase

Physics

1 answer:

SashulF [63]3 years ago

5 0

Answer:

E_Phase = 560V

Explanation:

The computation of the voltage i.e. dropped across each phase is shown below:

Given that

The delta connection line voltage is

E_line = 560 V

And, in the case of delta connection, the line voltage would be equivalent to the phase voltage

That means

E_Phase = E_Line

= 560 V

Hence, the voltage i.e. dropped across each phase is

E_Phase = 560V

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A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v

Pavel [41]

Answer:

Explanation:

The resonant frequencies for a fixed string is given by the formula nv/(2L).

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e fn+1 – fn = 390 -325

= 65

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3 years ago

A chemist needs to order an element that will not react with any other element. Which element should he order

Gwar [14]

Answer:

Helium

Explanation:

Helium is the least reactive element, since it is a noble gas with the smallest amount of valence rings.

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3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

A cart was pulled for a distance of 1 kilometer and the amount of work accomplished equaled 40,000 joules. With what force was t

goblinko [34]

Answer:

40 N

Explanation:

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3 years ago

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$