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3 years ago

What are the factors that affect the resistance of a wire?

Physics

1 answer:

777dan777 [17]3 years ago

6 0

1) Length of the wire.

2) Thickness of the wire.

3) Temperature.

4) Type of metal.

Hope this helps!

-Payshence

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What do scientitsts do to easily share mesurment data they can understand

Leona [35]

There are a few ways to do this- unfortunately different fields are better at it than others! Medical research is generally pretty good, some other fields likewise very good, some not as much. Basically, though, what they do is use standadisation- they agree on the terminology, units of data, statistical measures, and so forth, that will be used in that scientific field. As much as possible, every scientist in the field uses those standards so everyone working in the field should recognise it. For instance, in clinical trials, there is very good agreement worldwide on what the different metrics we use are- e.g. in cancer research, we usually want to know the 5-year survival rate (meaning the percentage of patients still alive 5 years after diagnosis). So anyone with the right training should be able to pick up a clinical trial report and understand what the results are and what the report is saying.

6 0

3 years ago

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0

3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$