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If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t

Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

μ₀ =permeability of free space 4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0

2 years ago

An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

$\displaystyle v_o=\sqrt{2gh_m}$

$v_o=\sqrt{2\cdot 9.8\cdot 77.5}$

$v_o=39\ m/s$

(b)

The maximum time or the time taken by the object to reach its highest point is calculated as follows:

$\displaystyle t_m=\frac{v_o}{g}$

$\displaystyle t_m=\frac{39}{9.8}$

$t_m=4\ s$

7 0

2 years ago

What part of the stem cell provides instructions for building the heart

Monica [59]

It’s DNA that’s a answer

3 0

2 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$