Part b what is the particle's speed at t=4.5s ? express your answer using two significant figures.

What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of

Murljashka [212]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

$A=\frac{\rho l}{R}\\ \frac{\pi d^2}{4}=\frac{\rho l}{R}\\ l=\frac{\pi d^2 R}{4 \rho}$

Here $d=0.15, R=15, \rho=1.68(10^{-8})$ .

Substituting numerical values,

$l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})} \\ l=15.78$

The length of the wire is $15.78 \;\;m$

5 0

4 years ago

A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

marin [14]

$a = \displaystyle\frac{v-u}{t}$

$-3 = \displaystyle\frac{v - 30}{3}$

$v -30 = -9$

$v = 21 m/s$

4 0

3 years ago

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What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

mr_godi [17]

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

$F=G*\frac{m1*m2}{r^{2} }$

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

$F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }$