All categories

4 years ago

A three-phase load has a phase voltage of 240 V and a phase current of 18 A. What is the apparent power of this load

Physics

1 answer:

Vlad1618 [11]4 years ago

6 0

Answer:

7482.5VA

Explanation:

We have that,

phase voltage of a three phase load is,240v

The phase current of a three phase load is, 18A

So To find apparent load we use

VA= √3*E*I

=√ 3*240*18

=7482.5VA

You might be interested in

Which of the following is NOT a type of model used in scientific inquiry? interactive, mathematical, conceptual, convergent

katrin2010 [14]

A conceptual model. Hope this helps.

5 0

3 years ago

Read 2 more answers

A 30kg mass is brought from the earth to the moon what is the weight on the earth

ozzi

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

8 0

2 years ago

6kg of human blood at a temperature of 65degees celcius is mixed with 4kg of human blood ata temperature of 20 degrees celcius .

Gennadij [26K]

Answer:

$T_f=47^{\circ}$

Explanation: Two samples of blood that have different masses and temperatures and are mixed, we have to find the final temperature of the mixture. the final temperature can be found using the following formula:

$T_f=\frac{(m_1\cdot T_1+m_2T_2)}{(m_1+m_2)}\Rightarrow(1)$

(1) Formula basically tells us that the product of mass and temperature remains constant throughout, so the addition of two products of the two separate blood samples would be equal to the product of final temperature and the total mass of the mixture. Mathematically this means that:

$\mleft(m_1+m_2\mright)T_f=(m_1\cdot T_1)+(m_2T_2)$

Using (1) and plugging in the corresponding values, we get the answer as follows:

$\begin{gathered} T_f=\frac{(m_1\cdot T_1+m_2T_2)}{(m_1+m_2)}\Rightarrow(1) \\ m_1=6\operatorname{kg} \\ m_2=4\operatorname{kg} \\ T_1=65^{\circ} \\ T_2=20^{\circ} \\ \therefore\rightarrow \\ T_f=\frac{(6kg\cdot65^{\circ}+4\operatorname{kg}\cdot20^{\circ})}{(6kg+4\operatorname{kg})}=\frac{(390+80)}{10}=\frac{470}{10}=47^{\circ} \\ \therefore\rightarrow \\ T_f=47^{\circ} \end{gathered}$

4 0

2 years ago

A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi

Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

Explanation:

(a)

2 Kg box causes tension

$T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N (b) Block mass of 4 Kg [tex]T'-mg sin \theta=0$ hence $T'=mg sin \theta$ where m is mass and g is gravitational force

T'=4*9.81 sin 35= 22.5071 N

Since T' is greater than $mg sin\theta$ , then the box moves down the slope

(c)

Acceleration a= $\frac {forward force-backward force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}$

$a= \frac {22.51-19.62}{2+4}=0.48$

When moving, the box will exert force T"= $mgsin \theta + ma$

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0

4 years ago

Read 2 more answers

If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru

Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

$\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}$

∴ Angle by which the second order line appear is 43.16°

5 0

3 years ago