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Rudik [331]
2 years ago
12

In a long straight wire, what current is required to exert a 1.0μN force on a 1.0μC charge moving at 1.5×106m/s parallel to the

wire at a distance of 0.35m from the wire?
Physics
1 answer:
Bas_tet [7]2 years ago
6 0

Answer:

Current in the wire is given as

i = 1.17 \times 10^{-7} A

Explanation:

magnetic field due to long current carrying wire is given as

B = \frac{\mu_0 i}{2\pi r}

so we have magnetic force on moving charge is given as

F = qvB

so we have

F = (1\times 10^{-6})(1.5 \times 10^6)(\frac{\mu_0 i}{2\pi (0.35)})

so we have

1\times 10^{-6} = 1.5 \times \frac{2 i}{0.35}

i = 1.17 \times 10^{-7} A

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At what time t is the turtle second time a distance of 10.0 cm from its starting point?
skad [1K]

Answer:

10 cm.

Explanation:

5 0
2 years ago
A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy
SSSSS [86.1K]

Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

\Delta U = m g \Delta h

where

m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta h= 2m is the change in height of the box

Substituting numbers, we find

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

3 0
2 years ago
A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
2 years ago
A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Irina18 [472]

Answer:

(b) 10 Wb

Explanation:

Given;

angle of inclination of magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ =  5.0 Wb

Magnetic flux is given as;

Φ = BACosθ

where;

B is the strength of magnetic field

A is the area of the plane

θ is the angle of inclination

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Now calculate the magnetic flux through a 2.0 m² portion of the same plane

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.

Option "b"

3 0
2 years ago
A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back
VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0
2 years ago
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