Determine one way you can contribute to water in the atmosphere in your day-to-day activities pleaseeeee helppp

An electric charge, A, is placed carefully between two other charges, B and C, and experiences no net electric force. Do B

Brrunno [24]

Answer:

I do not have enough information to tell

Explanation:

This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.

5 0

3 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

6 0

2 years ago

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago

A wire of radius 5 x 10⁻⁴ m is needed to prepare a coil of resistance 40 Ω. The resistivity of the material of the wire is 3.14x

SIZIF [17.4K]

Answer:

100 m

Explanation:

From the question,

R = Lρ/A.................... Equation 1

Where R = resistance of the wire, L = length of the wire, ρ = resistivity of the wire, A = cross sectional area of the wire.

But,

A = πr².................... Equation 2

Where r = radius of the wire.

Substitute equation 2 into equation 1

R = Lρ/πr²

Make L the subject of the equation

L = Rπr²/ρ...................... Equation 3

Given: R = 40 Ω, r = 5×10⁻⁴ m, ρ = 3.14×10⁻⁷ Ωm

Constant: π = 3.14

Substitute these values into equation 3

L = [40×3.14×( 5×10⁻⁴)²]/ (3.14×10⁻⁷)

L = 40×3.14×25×10⁻⁸/(3.14×10⁻⁷)

L = 100 m

Hence the length of the wire is 100 m

4 0

2 years ago

Examples of motor and generator?

Dafna11 [192]

Hope this helps

Answer:In a generator, current is produced in the armature winding. Ceiling fans, cars, etc. are all examples of motor. In power stations, generator is used to generate electricity.

6 0

2 years ago

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