What happens to the induced electric current if the number of loops is increased from one to three?

A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1

nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0

3 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?

pychu [463]

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

$H=ut+\frac{1}{2}gt^2$

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

$H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m$

3 0

3 years ago

Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more

masya89 [10]

Hey there!:

Here the Statement - D is correct.

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate. Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

6 0

3 years ago

Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less thanthat o

snow_tiger [21]

Answer:

41.4 g

41.4 cm³

1.08695 g/cm³

Explanation:

$\rho$ = Density of water = 1 g/cm³