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2 years ago

A transformer has a 240 V primary and a 60 V secondary. With a 5 ohm load connected, what is the primary current?

Physics

1 answer:

musickatia [10]2 years ago

4 0

Answer:primary current is 3A

Explanation:

primary voltage Vp=240V

secondary voltage Vs=60V

resistance 5Ω

primary current in the circuit is

$I_{P} =I_{S} \frac{V_{S}}{V_{P}}\\I_{P} =\frac{V_{S}}{R} \frac{V_{S}}{V_{P}}\\I_{P}=\frac{60\times 60}{5\times 240\\}\\I_{P}=3 A$

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A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emi

krok68 [10]

Answer:581.87 K

Explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

$\frac{4\pi}{3}r^3=a^3$

Surface area of sphere $A_s=4\pi r^2$

Surface area of cube $A_c=6a^2$

Total emmisive remains same

Thus $P=A\epsilon \sigma T^4$

$A_sT_s^4=A_cT_c^4$

$\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}$

$\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}$

$\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=614\times \frac{1.12679}{1.189}$

$T_c=581.87 K$

3 0

2 years ago

a 15kg television sits on a shelf at a height of 0.3 m how much gravitational potential energy is added to the television when i

DedPeter [7]

Answer:

<h2>103 Joules</h2>

Explanation:

In this problem we are required to find the potential energy possessed by the television

Given data

mass of television m = 15 kg

height added above the ground, h= 1-0.3 = 0.7 m

acceleration due to gravity g = 9.81 m/s^2

apply the formula for potential energy we have

P.E= m*g*h

P.E = 15*9.81*0.7 = 103 Joules

7 0

2 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

Even if all stars were the same distance from Earth, their absolute magnitude and

k0ka [10]

Answer: True

hope this helps!

4 0

1 year ago

Planet x has a mass of 4x1022 kg and a radius of 6x105 m What lise the grav ritational field strength in the surface of planet X

SCORPION-xisa [38]

Answer:

g ≈ 7.4 m/s²

Explanation:

The acceleration due to gravity on planet XX is ...

g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

g ≈ 7.4 m/s²

6 0

2 years ago