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3 years ago

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A wire is stretched right to its breaking point by a 5000 N force. A longer wire made of the same material has the same diameter

. Is the force that will stretch it right to its breaking point larger than, smaller than, or equal to 5000 N? Explain.

1 answer:

leva [86]3 years ago

4 0

Answer:

Equal to 5000N

Explanation:

The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.

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A single-turn square loop carries a current of 16 A . The loop is 15 cm on a side and has a mass of 3.8×10^−2kg . Initially the

DiKsa [7]

Answer:

The minimum magnetic field is 0.078 T.

Explanation:

Given that,

Current = 16 A

Side = 15 cm

Mass $m= 3.8\times10^{-2}\ kg$

Mass each segment in given square loop is

$m=\dfrac{3.8\times10^{-2}}{4}$

We need to calculate the torque due to gravity

Using formula of torque

$\tau_{g}=2mg(\dfrac{L}{2})+mgL$

$\tau_{g}=2mgL$

The torque due to magnetic field

$\tau_{B}=FL$

$\tau_{B}=BIL^2$

The equilibrium condition

$\tau_{B}=\tau_{g}$

Put the value into the formula

$BIL^2=2mgL$

$B=\dfrac{2mgL}{IL^2}$

$B=\dfrac{2mg}{IL}$

Put the value into the formula

$B=\dfrac{2\times\dfrac{3.8\times10^{-2}}{4}\times9.8}{16\times15\times10^{-2}}$

$B=0.078\ T$

Hence, The minimum magnetic field is 0.078 T.

7 0

3 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

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3 years ago

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Marina86 [1]

Hey! If this is on big ideas get a app called “Slader” It has answers to all math problems just like up the book you use > save it > type in the section you are doing > and look for the problems you are doing.

8 0

3 years ago

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

Anuta_ua [19.1K]

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

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3 years ago

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Sergio039 [100]

Answer:

I think D??

Explanation:

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