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RoseWind [281]
3 years ago
6

5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

nimize the power losses due to the high current heating up the line, a transformer steps up the potential difference to 500 000 V before it is transmitted. What is the current in the transmission lines?
Physics
1 answer:
lakkis [162]3 years ago
3 0

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

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A projectile is fired upward at an angle θ above the horizontal with an initial speed v0. At its maximum height, what are its ve
aliya0001 [1]

Answer:

\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y

Explanation:

The equations of kinematics will be used to solve this question:

y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt

At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.

First, let's apply the above equations to the x-direction.

There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.

v_x = v_{x_0} + a_xt = v_{x_0} + 0\\v_x = v_{x_0} = v\cos(\theta)

Therefore, the velocity vector of the projectile is

v_{max} = v_x = v\cos(\theta)

The speed of the projectile is the same.

The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.

7 0
4 years ago
the decimal reduction time (DRT) is the time it takes to kill 90% of cells present. Assume that a DRT value for autoclaving a cu
kotegsom [21]

Answer:

It takes 10.5 minutes to kill all the bacteria.

Only 1 cell would remain after 9 minutes.

Explanation:

It will take 1.5 minutes to kill 90% of the cells. So, after 1.5 minutes, only 10% would remain. After 3 minutes, only 1% remain. So, to figure out how long it would take to kill a million cells, we have to multiply 1 million by 0.1 repeatedly until the final value is less than 1 that is because when the value is less than 1, it means there are no more bacteria.

So:  

10^6 \times (0.1)^7 = 0.1  

So, you need 10.5 minutes of killing to kill one million cells.

Time taken=  7 x 1.5 minutes = 10.5 minutes.  

After 9 minutes you would have:  

10^{6} \times (0.1)^{6} = 1 cell left

6 0
4 years ago
If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt
Alik [6]

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

6 0
3 years ago
Read 2 more answers
Any one any kind of clue on this please need help
devlian [24]

I honestly think it is B, if not I'm sorry for the incorrect answer, but B seems to be the only one to make sense

6 0
3 years ago
We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What mu
Mama L [17]

Answer:

Explanation:

Given that the hoop makes one complete oscillation in 2s, this implies that the period of oscillation is 2s

Then,

T = 2s

Let Mass of the thin hoop be M

Let Radius of the hoop be R

Moment of inertial of a hoop is

I = MR²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

M is the mass of the hoop

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = R

Then, applying the formula

T = 2π √ (I / MgR)

So, we know that

I = MR²

Then,

T = 2π √( MR² / MgR)

T = 2π √(R/g)

Make R subject of formula

Square both sides

T² = 4π²(R/g)

T²g = 4π²R

Then, R = T²g / 4 π²

Since g = 9.8m/s² and T= 2s

R = 2² × 9.8 / 4π²

R = 0.993m

Then, the radius of the hoop is 0.993m

3 0
3 years ago
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