Ask question

Ask question

All categories

dimulka [17.4K]

3 years ago

7

Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t

hey are only one-eighth as far apart? Express your answer using two significant figures.

1 answer:

alexandr1967 [171]3 years ago

8 0

Answer:

New force, $F'=48\times 10^3\ N$

Explanation:

It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$ ..........(2)

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{1}{64}$

$F'=7.5\times 10^2\ N\times 64$

F' = 48000 N

or

$F'=48\times 10^3\ N$

Hence, this is the required solution.

You might be interested in

A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers

swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0

3 years ago

Read 2 more answers

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago

A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration

Luda [366]

Answer:

$a \approx \: 0.001 \: m {s}^{ - 2}$

Explanation:

Given:

$initial \: velocity \: (u) = 0 \\ \\ Final \: Velocity \: (v) = 72 km /h \\ \\ Time \: (t) = 5 \: hours \\ \\ Acceleration \: (a) =? \\ \\ \because \: a = \frac{v - u}{t} \\ \\ \therefore \: a = \frac{72 - 0}{5} \\ \\ a = \frac{72}{5} \\ \\ a = 14.5 \: km {h}^{ - 2} \\ \\ a = \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2} \\ \\ a = 0.00111882716 \\ \\ a \approx \: 0.001 \: m {s}^{ - 2}$

4 0

3 years ago

How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.

pav-90 [236]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

8 0

3 years ago

Charge A is sitting in an electric field you know the following information:________

soldi70 [24.7K]

Answer:

The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:

∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)

Explanation:

The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.

6 0

3 years ago