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2 years ago

I NEED HELP PLZZZZZZ

Physics

1 answer:

schepotkina [342]2 years ago

4 0

Mostly GPE and a little KE since the ball is high up (GPE) and it's also moving (KE) but not as much as it had when you first threw it

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current is passed through two parallel conductors in the same direction. If the conductors are placed near each other,they will

svlad2 [7]

If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will attract each other.

<h3>What is electric current?</h3>

Electric current can be defined as the flow of electrons.

Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.

Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.

On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.

The flow of current is known as electricity.

Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.

In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.

Learn more about parallel conductors at: brainly.com/question/17148082

#SPJ1

6 0

1 year ago

Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For

SpyIntel [72]

Answer:

$1.27\times 10^{12}\Omega/m$

Explanation:

We are given that

Diameter=d= $\mu m$

Thickness= $1\mu m$

Radius= $r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m$

Using $1\mu m=10^{-6} m$

Dielectric constant=8

Resistance = $R=2\times 10^5\Omega cm^2$

Internal specific resistance=r=100 ohm cm= $100\times \frac{1}{100}\Omega-m=1\Omega m$

Using 1 m=100 cm

Internal resistance per unit length= $\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m$

Using $\pi=3.14$

Internal resistance per unit length= $1.27\times 10^{12}\Omega/m$

8 0

2 years ago

PSYCHOLOGY <br> Which of these statements are false ?

nataly862011 [7]

Answer:

The false statement has to be that neruotransmitters are in the spinal cord.

Explanation:

Neruotransmitters are in no was associated with the spinal cord. That is more related to nerves and muscles. Neurotransmitters are in the Phasma Membrane acording to sciencedirect.com.

4 0

1 year ago

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

spacing between the slits is 405.32043 × $10^{-9}$ m

7 0

2 years ago

Observe the given figure and find the the gravitational force between m1 and m2.

Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

$F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}$

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

$F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N$

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0

2 years ago