atoms combine in different ways to make up all of the substances you encounter everyday true or false

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

4 0

2 years ago

Mention & reasons why the ability to adapt to change is

s2008m [1.1K]

Answer:

The ability to adapt is important because :

1) It helps in the survival of human beings.

2) It brings more variation to the human kind.

3) It helps the species from getting endangered or extinct.

4) It brings transformation in the adapting kind.

Hope this helps you☺️☺️

5 0

2 years ago

Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where

fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀ $e^{-at}$

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0

3 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

3 years ago

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

7 0

3 years ago