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Marrrta [24]
2 years ago
10

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

urrent is: A) 2.0 A B) 0.63 A C) 5.9A D) 300 A E) 26000 A
Physics
1 answer:
Mrac [35]2 years ago
7 0

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

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If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
2 years ago
A bicyclist travels the first 800 m of a trip 1.4 minutes, the next 500 m in 1.6 minutes, and finishes up the final 1200 m in 2
soldier1979 [14.2K]

Answer:

 v_average = 500 m / min

Explanation:

Average speed is defined

         v = (x_{f} -x₀) / Δt

let's look in each section

section 1

the variation of the distance is 800 in a time of 1.4 min

         v₁ = 800 / 1.4

         v₁ = 571.4 m / min

section 2

distance interval 500 in a 1.6 min time interval

         v₂ = 500 / 1.6

         v₂ = 312.5 m / min

section 3

distance interval 1200 m in a time 2 min

         v₃ = 1200/2

         v₃ = 600 m / min

taking the speed of each section we can calculate the average speed

         

the distance traveled

        Δx = 800 + 500 + 1200

        Δx = 2500 m

the time spent

        Δt = 1.4 + 1.6+ 2

        Δt = 5 min

         v_average = Δx / Δt

         v_average = 2500/5

         v_average = 500 m / min

7 0
2 years ago
In the diagram below, what is the property of the wave indicated by the letter A? a.Crest
Ugo [173]
Do you have a picture of the diagram that I could view?
4 0
2 years ago
Read 2 more answers
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
1 year ago
When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?
Svetach [21]
When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
and
-- that's the lowest it's going to get, so its potential energy is minimum.

'c' is your choice.
5 0
2 years ago
Read 2 more answers
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