Ask question

Ask question

All categories

1 year ago

10

Suppose you wish to fabricate a uniform wire from 1.00 g of copper. If the wire is to have a resistance of R= 0.500ω and all the

copper is to be used, what must be(a) the length and

1 answer:

Alekssandra [29.7K]1 year ago

3 0

In order to fabricate a uniform wire that has a resistance of R = 0.500Ω from 1.00 g of copper and all the copper is to be used, the Length, L of the copper required is 1.82m.

<h3>What would be the length of the copper wire?</h3>

The length of the wire is obtained as follows:

Length, L = √R*m/e*d

where;

R is resistance and is equal to 0.50 Ω

m is mass of the copper and is equal to 1.00 g or 10⁻³ kg

e, resistivity of Copper = 1.7 x 10⁻⁸ Ω⁻¹m

density of copper, d = 8.92 x 10³ kg/m³

L = √(10⁻³ * 0.50/1.7 x 10⁻⁸ * 8.92 x 10³)

Length, L= 1.82m

In conclusion, the length of the wire is obtained from the mass and density of the copper wire.

Learn more about length of wire at: https://brainly.in/question/49182286

#SPJ4

You might be interested in

If the angle of refraction is 20 degrees what is the angle of incidence

Romashka-Z-Leto [24]

Answer:

13.1

Explanation:

that's what I'm gonna go with, but u can research more

5 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0

3 years ago

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0

4 years ago

A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp

alex41 [277]

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which states that:

$\sum F=ma$

where:

$\sum F$ is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

$\sum F=30 N - 20 N = 10 N$ (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

$a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2$

in the same direction as the force (left).

5 0

3 years ago

If density of water is 300kg m3 and hight 4m of water find the pressure of water ?<br>

iren [92.7K]

Answer:

11, 760 Pa.

Explanation:

By applying formula P= pgh, where P is pressure, p is density, g is gravitational acceleration (9.8 m\s2) and h is height of water level. Putting values in the formula, you can have the correct answer.

3 0

3 years ago