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2 years ago

What is meant by the statement,the linear expansivity of copper is 0.000017k

Physics

1 answer:

Aliun [14]2 years ago

4 0

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

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Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

7 0

3 years ago

An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases

egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0

3 years ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago

Scientists have proven that genes play no role in self-esteem. Please select the best answer from the choices provided. T F

IRINA_888 [86]

The answer is False give thanks for the answer m8 and happy Halloween

6 0

3 years ago

Read 2 more answers