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3 years ago

How can the potential energy in magnets be used to create kinetic energy in an object without making contact?

1 answer:

7 0

Answer: If you move a magnet quickly through a coil of copper wire, the electrons will move - this produces electricity.

Explanation:

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Which particle is most likely to interact with your hand? Select one: a. Alpha particle b. Beta particle c. Gamma particle d. Ne

IgorLugansk [536]

Answer:

The correct option is a

Explanation:

The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).

From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.

4 0

3 years ago

If there is no electric current flowing through a wire then what happens to the magnetic field

Zinaida [17]

If there is no current in the wire .....the direction of magnetic field remains unchanged

7 0

2 years ago

Read 2 more answers

A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,

Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

$v_f^2=v_o^2+2a*d$ Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

$v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}$

$v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}$

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

$15.4^2 = 30.8^2+2(-0.5)*d$

$2(0.5)*d = 30.8^2 - 15.4^2$

$d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m$

$d = 717.6 m\frac{1km}{1000m} = 0.7176Km$

8 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC

x₁ = - 1.715m x₃ = - 1.085m x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0

3 years ago

A wire loop with 70 turns is formed into a square with sides of length ???? . The loop is in the presence of a 1.20 T uniform ma

Oksi-84 [34.3K]

Answer:

l= 3.002 cm

Explanation:

Given that

n= 70 turns

B= 1.2 T

θ= 15°

I= 1.5 A

τ = 0.0294 N⋅m

Lets take length of sides is l.

We know that

τ = n I A B sin θ

Area of square ,A= l²

Now by putting the value

τ = n I A B sin θ

0.0294 = 70 x 1.5 x l² x 1.2 x sin 15°

l² = 0.000901 m²

l² = 9.01 cm²

l= 3.002 cm

3 0

3 years ago