A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) is the force attractive or repulsive?

Answer:

a) 2.24°

b) 11.99 km

c) 0.9992 miles

Explanation:

We can think of the road as a triangle with a 12 km hypotenuse and a 0.5 km side. Then:

l = 12

h = 0.5

and

sin(a) = h / l

a = arcsin(h / l)

a = arcsin(0.5 / 12)

a = 2.24°

That is the slope.

The map distance travelled would be the other side of the triange.

cos(a) = d / l

d = l* cos(a)

d = 12 * cos(2.24) = 11.99 km

And for miles

d = 1 mile * cos(2.24) = 0.9992 miles

Answer:

If an object moves twice as fast its kinetic energy quadruples.

Explanation:

The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:

K₁= 1/2 * m* v²

If we double the speed of the body, its kinetic energy (K₂) will be:

K₂= 1/2 * m*( 2v)²

K₂= 1/2 * m* 4 *v²

K₂= 4(1/2 * m *v²)

K₂= 4*K₁

Answer:

r=2.6 cm

Explanation:

Hi!

Lets call steel material 1, and the new alloy material 2. You know their densities:

[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]

The volume of a sphere with radius r is given by:

[tex]V=\frac{4}{3}\pi r^3[/tex]

Then the masses of the bearings are:

[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]

For the masses to be the same:

[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]

Answer:

(A) - 0.273 kg m /s

(B) - 0.546 kg m /s

Explanation:

mass of paintball, m = 0.0032 kg

initial velocity, u = 85.3 m/s

(A) Momentum is defined as the product of mass of body and the velocity of body.

initial momentum, pi = mass x initial velocity

pi = 0.0032 x 85.3 = 0.273 kg m /s

Finally the ball stops, so final velocity, v = 0 m/s

final momentum, pf = mass x final velocity = 0.0032 x 0 = 0 m/s

change in momentum = pf - pi = 0 - 0.273 = - 0.273 kg m /s

(B) initial velocity, u = + 85.3 m/s

final velocity, v = - 85.3 m/s

initial momentum, pi = mass x initial velocity

pi = 0.0032 x 85.3 = 0.273 kg m /s

final momentum, pf = mass x final velocity = - 0.0032 x 85.3 = - 0.273 kgm/s

change in momentum = pf - pi = - 0.273 - 0.273 = - 0.546 kg m /s

(C) According to the Newton's second law, the rate of change of momentum is directly proportional to the force exerted.

As the change in momentum in case (B) is more than the change in momentum in case (A), so the force exerted on the skin is more in case (B).

Answer:

The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Solution:

As per the question:

[tex]v_{1} = 3 x 10^5 m/s[/tex]

[tex]v_{2} = 6.40 x 10^6 m/s[/tex]  

length of the re region, l = 1 cm = 0.01 m

Now,

Using the third equation of motion:

[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]  

[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]  

[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]  

[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at²    .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m

Answer:

Electric field at a distance of 1.45 cm will be [tex]172.41\times 10^4N/C[/tex]

Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

And the potential difference [tex]V=2.5\times 10^4volt[/tex]

There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential difference and d is distance

So [tex]E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C[/tex]

The magnitude of the uniform electric field between the plates is approximately [tex]\( 1.724 \times 10^4 \) V/m.[/tex]

The magnitude of the uniform electric field between the plates is [tex]\( E = \frac{\Delta V}{d} \)[/tex], where [tex]\( \Delta V \)[/tex] is the potential difference and ( d ) is the distance between the plates.

Now, we can calculate the electric field ( E ):

[tex]\[ E = \frac{\Delta V}{d} = \frac{2.50 \times 10^4 \text{ V}}{1.45 \times 10^{-2} \text{ m}} \][/tex]

[tex]\[ E = \frac{2.50 \times 10^4}{1.45} \times 10^2 \text{ V/m} \][/tex]

[tex]\[ E = 1.724 \times 10^4 \text{ V/m} \][/tex]

Answer:

t = 0.018 s  

Explanation:

given,

nerve impulse in human body travel at a speed of  = 100 m/s

height of the man = 1.80 m

time taken by the impulse to travel from foot to brain = ?

distance = speed × time                      

1.80 = 100 × t                                              

t =[tex]\dfrac{1.80}{100}[/tex]                                    

t = 0.018 s                            

hence, the time taken by the nerve to reach brain from toe is 0.018 s

Answer:

9 m/s

Explanation:

mass of cannon, M = 100 kg

mass of cannon ball, m = 10 kg

velocity of cannon ball, v = 90 m/s

Let the recoil velocity of cannon is V.

Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.

Momentum before the firing = momentum after the firing

M x 0 + m x 0 = M x V + m x v

0 = 100 x V + 10 x 90

V = - 9 m/s

Thus, the recoil velocity of cannon is 9 m/s.

The pebble's initial speed is approximately 22 m/s. It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.

To determine the pebble's initial speed, we can use the equation for projectile motion:

h = v0yt - (1/2)gt2

Where h is the height, v0y is the initial vertical speed, t is the time, and g is the acceleration due to gravity.

Since the pebble is launched straight up, the final height is equal to the initial height. Plugging in the given values, we have:

37 m = v0y(2.3 s) - (1/2)(9.8 m/s2)(2.3 s)2

Simplifying this equation gives us the value of v0y, the initial vertical speed. To find the pebble's initial speed, we can use the Pythagorean theorem:

v0 = √(v0x2 + v0y2)

Where v0 is the initial speed and v0x is the initial horizontal speed. Since the pebble is launched straight up, v0x = 0. Plugging in the calculated value of v0y, we can solve for v0.

(a) The pebble's initial speed is approximately 22 m/s.

(b) To find the time it takes for the pebble to first reach a height of 18.5 m, we can use the equation for height:

18.5 m = v0yt - (1/2)gt2

Solving for t gives the time it takes for the pebble to reach the desired height.

(b) It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.

Answer:

Explanation:

The force [tex]\vec{F}[/tex] on a charge q made by an electric field [tex]\vec{E}[/tex] its

[tex]\vec{F} = q \vec{E}[/tex]

The electric charge of the electron its

[tex]q \ = \ - \ 1.602 \ 10 ^{-19} \ C[/tex].

Taking the unit vector [tex]\hat{i}[/tex] pointing towards the east, the electric field will be:

[tex]\vec{E}= 3400 \ \frac{N}{C} \ \hat{i}[/tex].

So, the force will be:

[tex]\vec{F} =  \ - \ 1.602 \ 10 ^{-19} \ C \ * \ 3400 \ \frac{N}{C} \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5446.8 \ 10 ^{-19} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.

Answer:

1.03×10³ kg/m³

Explanation:

β = Bulk modulus = 2.34×10⁹ N/m²

f = Frequency = 1000 Hz

λ = Wavelength = 1.51 m

ρ = Density

Speed of the wave

v = fλ

⇒v = 1000×1.51

⇒v = 1510 m/s

Speed of sound

[tex]v=\sqrt{\frac{\beta}{\rho}}\\\Rightarrow \rho=\frac{\beta}{v^2}\\\Rightarrow \rho=\frac{2.34\times 10^9}{1510^2}\\\Rightarrow v=1026.27\ kg/m^3[/tex]

Density of seawater is 1.03×10³ kg/m³ or 1.03 g/cm³ (rounding)

Answer:

a) vertical position at 2s: 50.68 m

b) horizontal position at 2s: 13.05 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Where:

[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed

[tex]\theta=25\°[/tex] is the angle at which the rock was thrown

[tex]t=2 s[/tex] is the time

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (2)

Where:

[tex]y_{o}=25 m[/tex]  is the initial height of the rock

[tex]y[/tex]  is the height of the rock at 2 s

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin with the answers:

a) Vertical position at 2 s:

[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex]   (3)

[tex]y=25 m+6.085 m-19. 6 m[/tex]   (4)

[tex]y=50.68 m[/tex]   (5)  This is the vertical position at 2 s

b) Horizontal position at 2 s:

[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex]    (5)

[tex]x=13.05 m[/tex]   (6)  This is the horizontal position at 2 s

Explanation:

It is given that,

Position of the man, h = 25 m

Initial velocity of the rock, u = 7.2 m/s

The rock is thrown upward at an angle of 25 degrees.

(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :

[tex]y=-\dfrac{1}{2}gt^2+u\ sin\theta\ t+h[/tex]

[tex]y=-\dfrac{1}{2}\times 9.8(2)^2+7.2\ sin(25)\times 2+25[/tex]

y = 11.48 meters

(b) Let x is the horizontal position of the rock at that time. It can be calculated as :

[tex]x=u\ cos\theta \times t[/tex]

[tex]x=7.2\ cos(25)\times 2[/tex]

x = 13.05 meters

Hence, this is the required solution.

Answer:

W = 0

Explanation:

given data:

starting position is ( 2cm , 7 cm)

ending position is  (10 cm, 7 cm)

we know thta work done is given as

[tex] W = q\Delta v[/tex]

here , q is particle charge

[tex]\Delta v[/tex] is change in potential

as we know that intial and final position are having uniform eleectric potential, therefore it have same potential i.e. V_1 = V_2, thus

one more reason of being work done equal to zero is,

Since the distance covered by the item and the direction of motion is always perpendicular to each other during the round journey, no work is therefore carried out.

W = 0

Answer:

The charge at the center of the conducting sphere is zero.

Explanation:

A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.

Answer:1.624 m

Explanation:

Given

height of 2nd floor =5 m

time recorded is 0.58 sec

Let h be the height of third floor above 2 nd floor

[tex]h=ut+\frac{at^2}{2}[/tex]

here u=0

t=time taken to cover height h

[tex]h=\frac{gt^2}{2}[/tex]   ----------1

Now time taken to complete whole building length

[tex]h+5=\frac{g(t+0.58)^2}{2}[/tex]  ----------2

Subtract 1 form 2

[tex]5=\frac{g(2t+0.58)(0.58)}{2}[/tex]

Taking gravity [tex]g=1 m/s^2[/tex]

1=(2t+0.58)(0.58)

t=0.57 s

Substitute the value of t in 1

h=1.624 m

Answer:

option C

Explanation:

the correct answer is option C.

Graph between velocity time which shows the constant velocity is straight line  with slope zero.

constant velocity means velocity is not changing with respect to time hence this condition will be only when graph is straight line with slope zero.

area under velocity time graph shows the displacement of the object.

And where as slope of velocity time graph show acceleration of the object.

Final answer:

The correct option is (d) A straight line sloping upward. A graph of velocity vs time that shows an object moving toward a motion detector at a constant speed is represented as a straight line sloping upward, indicating a constant positive velocity.

Explanation:

When considering an object moving toward a motion detector at a constant speed, the velocity vs. time graph should depict a constant velocity. This is visualized as a straight line on such a graph, since the speed of the object does not change over time.

The correct choice to represent an object moving at a constant speed towards a motion detector is (d) A straight line sloping upward.

The key concept here is that when velocity is constant, the slope of the velocity vs. time graph remains constant. If the object is moving toward the motion detector, the velocity is positive, therefore, the graph slopes upwards, indicating a positive velocity that does not change.

This scenario clearly shows the relationship between velocity and time when an object moves at a constant speed.

Answer:

The second statement is true: If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.

Explanation:

Hi there!

Let´s write the rate law for the original reaction and the reaction with X increased by 1.5:

rate 1 =k [X][Y]²

rate 2 = k[1.5 X][Y]²

Now we have to demonstrate if rate 2 = 2.25 rate 1

Let´s do the cocient between the two rates:

rate 2/ rate 1

if rate 2 = 2.25 rate 1

Then,

rate 2 / rate 1 = 2.25 rate 1 / rate 1 = 2.25

Let´s see if this is true using the expressions for the rate law:

rate 2 / rate 1

k[1.5 X][Y]² / k [X][Y]² = 1.5 k [X][Y]² / k[X][Y]² = 1.5

2.25 ≠ 1.5

Then the first statement is false.

Now let´s write the two expressions of the rate law, but this time Y will be increased by 1.5:

rate 1 = k[X][Y]²

rate 2 = k[X][1.5Y]²

Again let´s divide both expressions to see if the result is 2.25

rate 2 / rate 1

k[X][1.5Y]²/ k [X][Y]²  

(distributing the exponent)

(1.5)²k [X][Y]² / k [X][Y]² = (1.5)² = 2.25

Then the second statement is true!

Answer:

a) True

b)False

c) True

Explanation:

The order of reactants decides the exponents of respective reactant concentrations.

Since X is first order, exponent is = 1

Overall third order, Exponent X + Exponent Y = 3

Hence, exponent Y = 2.

Hence,

Rate = k[X][Y]^2

b)

If X conc is increased by 1.5 Rate should increase by 1.5 because k proportional to [X]. Hence, statement is False

c)

If Y conc is increased by 1.5 Rate should increase by 2.25 because k proportional to [Y]^2 = (1.5)^2 = 2.25. Hence, statement is True

Answer:

The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).

If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

[tex]V=-2K\lambda ln\left (\frac{r'}{r}  \right )[/tex]

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

[tex]qV = 0.5m(u^{2}-v^{2})[/tex]

By substituting the values, we get

[tex]V=\frac{mu^{2}}{2q}[/tex]

[tex]-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=0.35[/tex]

[tex]\frac{r'}{r} =e^{-0.35}[/tex]

[tex]\frac{r'}{r} =0.7047[/tex]

r' = 14.5 x 0.7047 = 10.22 cm

(a) V = [tex]\frac{8.3}{P}[/tex]

(b) (i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]

(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.

(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]

Boyle's law states that at constant temperature;

P ∝ 1 / V

=> P = k / V

=> PV = k   -------------------------(i)

Where;

P = pressure

V = volume

k = constant of proportionality

According to the question;

When;

V = 0.106m³, P = 50kPa

Substitute these values into equation (i) as follows;

50 x 0.106 = k

Solve for k;

k = 5.3 kPa m³

(a) To write V as a function of P, substitute the value of k into equation (i) as follows;

PV = k

PV = 8.3

Make V subject of the formula in the above equation as follows;

V = 8.3/P

=> V = [tex]\frac{8.3}{P}[/tex]        -------------------(ii)

(b) Find the derivative of equation (ii)  with respect to V to get dV/dP as follows;

V = [tex]\frac{8.3}{P}[/tex]

V = 8.3P⁻¹

[tex]\frac{dV}{dP}[/tex] = -8.3P⁻²

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{P^{2} }[/tex]

Now substitute P = 50kPa into the equation as follows;

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{50^{2} }[/tex]   [[tex]\frac{kPam^{3} }{(kPa)^{2} }[/tex]]               -----  Write and evaluate the units alongside

[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{2500 }[/tex]    [[tex]\frac{kPam^{3} }{k^{2} Pa^{2} }[/tex]]

[tex]\frac{dV}{dP}[/tex] = - 0.00332 [[tex]\frac{m^{3} }{kPa}[/tex]]

Therefore,

(i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]

(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.

(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]

Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. To express volume as a function of pressure, use the formula V = k/P. The derivative of volume with respect to pressure indicates the rate at which volume changes as pressure changes, measured in cubic meters per kilopascal (m3/kPa).

Boyle’s Law states that for a given amount of gas at a constant temperature, the volume V is inversely proportional to the pressure P. This relationship can be mathematically represented as PV = k, where k is a constant.

(a) Given a sample of air at 25°C with a volume of 0.106 m3 and a pressure of 50 kPa, we can write the volume as a function of pressure by rearranging the formula to V = k/P. Using the initial conditions, we find that k = P × V = 50 kPa × 0.106 m3 = 5.3 kPa · m3, so V(P) = 5.3 kPa · m3 / P.

(b) To calculate dV/dP when P = 50 kPa, we differentiate the function V(P) with respect to P, resulting in dV/dP = -5.3 kPa · m3 / P2. At P = 50 kPa, this becomes dV/dP = -5.3 / (502) = -0.00212 m3/kPa. The derivative represents the rate of change of volume with respect to pressure, which in this case means that for each increase of 1 kPa in pressure, the volume decreases by 0.00212 m3. The units of the derivative are m3/kPa.

Answer:

(a) 1000 N/C

Explanation:

Kinetic energy of electron, K = 1.6 x 10^-17 J

distance, d = 10 cm = 0.1 m

Let the potential difference is V and the electric field is E.

(a) The relation between the kinetic energy and the potential difference is

K = e V

V = K / e

Where, e be the electronic charge = 1.6 x 10^-19 C

V = [tex]\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}[/tex]

V = 100 V

The relation between the electric field and the potential difference is given by

V = E x d

100 = E x 0.1

E = 1000 N/C

(b) The force acting on the electron, F = q E

where q be the charge on electron

So, F = -e x E

It means the direction of electric field and the force are both opposite to each other.

The direction of electric field and the force on electron is shown in the diagram.

Answer:

[tex]N=1.375*10^{11}[/tex] electrons

Explanation:

The total charge Q+ at the coin is equal, but with opposite sign, to the charge negative Q- removed of it. Q- is the sum of the charge of the N electrons removed from the coin:

[tex]Q_{-}=N*q_{e}[/tex]

[tex]Q_{-}=-Q_{+}[/tex]

q_{e}=-1.6*10^{-19}C    charge of a electron

We solve to find N:

[tex]N=-Q_{+}/q_{e}=-2.2*10^{-8}/(-1.6*10^{-19})=1.375*10^{11}[/tex]

Answer:

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

Explanation:

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

Force between them F = 8 X 10⁻³ N distance between them d = 4.1 m

F = k ( q₁ xq₂ ) / d²

8 x 10⁻³ = [tex]\frac{8.99\times10^9\times q_2\times (10^{-5}-q_2)}{(4.1)^2}[/tex]

14.95 x 10⁻¹² = 10⁻⁵q₂ - q₂²

q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹² =0

This is a quadratic equation .The solution of it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

q₁ = 3.2 x 10⁻⁶ C

q₂ = 6.8 x 10⁻⁶C

Charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.

It is also known as electric charge, electrical charge, or electrostatic charge and symbolized q.

Given,

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

F = 8 X 10⁻³ N

d = 4.1 m

By using the Formula,

[tex]F =\frac{k ( q_{1} * q_{2} ) }{d^{2} }[/tex]

8 x 10⁻³ = [tex]\frac{8.99*10^{9}*q_{2} *10^{-5} -q_{2} }{(4.1)^{2} }[/tex]

14.95 x 10⁻¹²

= 10⁻⁵q₂ - q₂²

= q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹²=0

Therefore,

The above equation is a quadratic equation , it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

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Answer:

(c) The forces have the same magnitude.

Explanation:

The electric Force between the two charges are proportional to the product of both charges:

[tex]F_{electric}=k*q_{1}q_{2}/r^{2}[/tex]

Both charges feels the same force but with opposite direction.

(c) The forces have the same magnitude.

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east  is = x

so distance from starting point = 5.21 × 10³ - x        ...................1

and we can say time required to reach end

time required = distance / speed

time required = [tex]\frac{5.21 *10^3 - x}{1.32}[/tex]    ................2

and

time required for 6.44 km west

time required = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]    ................3

and time required for distance x

time required = [tex]\frac{x}{0.687}[/tex]    ................4

so from equation 2 , 3 and 4

[tex]\frac{5.21 *10^3 - x}{1.32}[/tex] = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]  + [tex]\frac{x}{0.687}[/tex]

x = 1273.78 m

so distance in east is 1273.78 m

Answer:

The depth of well equals 120.47 meters.

Explanation:

The time it takes the sound to reach our ears is the sum of:

1) Time taken by the rock to reach the well floor.

2) Time taken by the sound to reach our ears.

The Time taken by the rock to reach the well floor can be calculated using second equation of kinematics as:

  [tex]h=ut+\frac{1}{2}gt^2[/tex]

where,

'u' is initial velocity

't' is the time to cover a distance 'h' which in our case shall be the depth of well.

'g' is acceleration due to gravity

Since the rock is dropped from rest hence we infer that the initial velocity of the rock =0 m/s.

Thus the time to reach the well base equals

[tex]h=\frac{1}{2}gt_1^2\\\\\therefore t_{1}=\sqrt\frac{2h}{g}[/tex]  

Since the propagation of the sound back to our ears takes place at constant speed hence the time taken in the part 2 is calculated as

[tex]t_{2}=\frac{h}{Speed}=\frac{h}{340}[/tex]

Now since it is given that

[tex]t_{1}+t_{2}=5.50[/tex]

Upon solving we get

[tex]\sqrt{\frac{2h}{g}}+\frac{h}{340}=5.5\\\\\sqrt{\frac{2h}{g}}=5.5-\frac{h}{340}\\\\(\sqrt{\frac{2h}{g}})^{2}=(5.5-\frac{h}{340})^{2}\\\\\frac{2h}{g}=5.5^2+\frac{h^2}{(340^2)}-2\times 5.5\times \frac{h}{340}[/tex]

Solving the quadratic equation for 'h' we get

h = 120.47 meters.

Answer:

a) The average velocity is 32 m/s

b) See the attached figure. Slope of the line = 32.

Graphic: a line that passes through the points (0; 50) and (5; 210)

Explanation:

a) The average velocity is calculated as the displacement over time:

v = ΔX / Δt

where

ΔX : displacement ( final position - initial position)

Δt : time (final time - initial time)

Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

Δt  = 5 s

v = 160 m / 5 s = 32 m/s

b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required [tex]= \frac{(14.644 x 10^6)}{784}[/tex]

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

Answer:

[tex]F=1.07\times 10^{-3}\ N[/tex]

Explanation:

Given that,

Charge on two balls, [tex]q_1=q_2=-38\ nC=-38\times 10^{-9}\ C[/tex]

Distance between charges, r = 11 cm = 0.11 m

We need to find the repulsive force between balls. Mathematically, it is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(38\times 10^{-9})^2}{(0.11)^2}[/tex]

F = 0.001074 N

or

[tex]F=1.07\times 10^{-3}\ N[/tex]

So, the magnitude of repulsive force between the balls is [tex]1.07\times 10^{-3}\ N[/tex].