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3 years ago

The magnetic field a distance 2 cm from a long straight current-carrying wire is 2 × 10–5 t. the current in the wire is:

Physics

1 answer:

Sliva [168]3 years ago

6 0

Current in the wire = 2 A

Explanation:

the magnetic field is given by

B= \frac{\mu i}{2\pi r}

μo= 4π x 10⁻⁷ Tm/A

i= current

r=0.02 m

B = magnetic field= 2 x 10⁻⁵ T

2 x 10⁻⁵= (4π x 10⁻⁷)(i) / (2π*0.02)

i=2 A

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The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.

dedylja [7]

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

Voltage = 230V

Power = 1500W

To find the resistance, R;

Power = V²/R

Where:

V is the voltage measured in volts.

R is the resistance measured in ohms.

Substituting into the equation, we have;

1500 = 230²/R

Cross-multiplying, we have;

1500R = 52900

R = 52900/1500

R = 35.27 Ohms.

Therefore, the resistance which the heating element needs to have is 35.27 Ohms.

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3 years ago

True or false gravitational pull decreases with an increase of distance between two objects

ohaa [14]

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

5 0

3 years ago

Which scenarios are examples of physical changes? Select three options

pashok25 [27]

Answer:

A kid becoming an adult

A leg becoming bruised

A person's blood pressure raising because they are running

need a picture to answer specific questions.

Explanation:

All of these are physical changes. Hope that this helps you and have a great day :)

4 0

3 years ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$