All categories

3 years ago

Can someone solve this problem and explain to me how you got it

Physics

1 answer:

likoan [24]3 years ago

3 0

Answer:

Explanation:

Coulomb's law states that the force of attraction or repulsion between any two charges is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between the charges

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question 7:

Given: $F=1.9*10^{-29}N$ , $q1=q2=1.6*10^{-19}C$

By coulomb's law,

$1.9*10^{-29}=9*10^{9}*\frac{1.6*10^{-19}*1.6*10^{-19}}{r^{2}}$

⇒ $r=3.5m$

Question 6:

Given: $q1=q2=-1.5*10^{-6}C$ , $r=0.28m$

By coulomb's law,

$F=9*10^{9}*\frac{1.5*10^{-6}*1.5*10^{-6}}{0.28^{2}}$

⇒ $F=0.26N$

You might be interested in

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the o

bezimeni [28]

Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question

8 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

crimeas [40]

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

Solving the equations gives:

$F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}$

$F_R = 238N\\ F_N = 1310N$

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

c) Using the result from b and solving for $r_F$

$\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m$

4 0

3 years ago

Soil formation involves both

Softa [21]

The process of soil formation involves the following steps: Accumulation of materials, leaching and losses (because of weather factors particles are leached and eroded away or taken up from the soil by plants), Transformation and illluviation (the chemical weathering of silt, sand, and the formation of clay minerals as well as the change of organic materials into decay resistant organic matter), podsolisation and translocations (strong acidic solutions breakdown the clay minerals). From the given options soil formation involves both physical and chemical weathering of rocks. Correct answer:D

6 0

2 years ago

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

The direction of motion of the person relative to the water is 16.7° north of east.

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$