JUDET

A loop of wire is placed in a magnetic field such that it has a flux, LaTeX: \phi ϕ, through it. The loop is compressed so that

Juliette [100K]

Answer:

B'= 3.333 B

Explanation:

Lets take

Initial area = A

Magnetic field = B

The area after compression

A'=0.3 A

Magnetic field = B'

We know that flux ,Ф

Ф = B A

Given that flux is constant so

B A = B' A'

B A=B' x 0.3 A

B'= 3.333 B

It means that magnetic field will increase.

6 0

3 years ago

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Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles

Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

The right answer is C)

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0

3 years ago

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How does the motion of the molecules change when you increase the temperature

borishaifa [10]

Answer:

They speed up as temperature increases.

Explanation:

As temperatures increases within a molecule the particles will speed up.

6 0

3 years ago

A crane operater drops the cargo by turning off the electric current to the

NikAS [45]

Answer:

A crane operator drops the cargo by turning off the electric current to the electromagnet. Electromagnets can be controlled by controlling the electric current. An electromagnet is stronger if there is more current flowing through it.

Explanation:

7 0

2 years ago

A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reac

Step2247 [10]

Answer:

$h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}$

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t Equation (1)

Where:

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀ : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y² $h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}$

2gh= (v₀*sin θ₀)²

$h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}$

7 0

3 years ago