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Neko [114]
3 years ago
12

A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and t

he secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.
Physics
1 answer:
Zina [86]3 years ago
3 0

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

\frac{V_1}{N_1}=\frac{V_2}{N_2}

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

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kirza4 [7]

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

F=T=\frac{mv^2}{L}

v=\sqrt{\frac{FL}{m}}

8 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
Have the highest birth rate
Sveta_85 [38]

Answer:

353225

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3 0
3 years ago
25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression
levacccp [35]

maximum speed of cheetah is

v_1 = v_{max}

speed of gazelle is given as

v_2 = v_{g}

Now the relative speed of Cheetah with respect to Gazelle

v_{12} = v_1 - v_2

v_{12} = v_{max} - v_g

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

d = v_{12}* t

so by rearranging the terms we can say

t = \frac{d}{v_{12}}

t = \frac{d}{v_{max} - v_g}

so above is the relation between all given variable

6 0
3 years ago
A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without
kari74 [83]

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

7 0
3 years ago
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