50 km equals how many meters?

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:

(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.

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:

[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:

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]

The temperature of the boiling chlorine is calculated using the resistance change of a platinum wire thermometer and the temperature coefficient of resistivity. After calculations, the boiling point of chlorine is found to be -34.6°C.

The temperature of the boiling chlorine can be deduced using the relation between temperature change and resistance change for a platinum wire thermometer. The equation we use is:

R = R0 (1 + αΔT)

Where R represents the resistance at the new temperature, R0 the resistance at a reference temperature (20°C in this case), α the temperature coefficient of resistivity, and ΔT the change in temperature. We can rearrange this equation to solve for the new temperature:

ΔT = ∂(T₂ - T₁)

Giving us:

(99.6 ohms / 125 ohms - 1) = 3.72 × 10-3°C-1 ∂T

Simplifying:

ΔT = (0.7968 - 1) / 3.72 × 10-3°C-1

ΔT = -0.2032 / 3.72 × 10-3°C-1

ΔT = -54.6°C

Thus, the boiling point of chlorine is:

20°C – 54.6°C = -34.6°C

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]

(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:

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: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:

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:

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:

the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt

Explanation:

We have given charge [tex]q=4\mu C=4\times 10^{-6}C[/tex]

Distance between the charge r = 16 cm = 0.16 m

Electric potential is given by [tex]V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}[/tex], here K  is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So potential [tex]V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt[/tex]

So the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt

The electric potential, V, of a point charge, is found using the formula V = kq/r. Given the charge of 4.00 μC and the distance of 16.0 cm, the electric potential comes out to approximately 2.24 x 10⁵ V.

The electric potential, V, of a point charge, can be calculated by the formula V = kq/r where:

In the question, q is given as 4.00 μC, which is equivalent to 4.00 x 10⁻⁶ C. The distance r is given as 16.0 cm, which is equivalent to 0.16 m. We get by substituting the values into the formula:

V = (8.99 × 10⁹ Nm²/C² x 4.00 x 10⁻⁶ C)/0.16 m

Through calculation, the electric potential 16.0 cm from a 4.00 μC point charge is therefore approximately 2.24 x 10⁵ V.

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Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the position must also be changing.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.

The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.

The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.

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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:

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:

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:

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:

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:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = [tex]\frac{3}{5}\times 15 = 9[/tex]

So the lines terminating at - 2 micro coulomb

                                    = [tex]\frac{2}{5}\times 15 = 6[/tex]

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

The number of lines emanating from the negative charged particles are 9.09 and 6.06 respectively.

In an electric field, the number of lines emanating from a charged particle is directly proportional to the magnitude of the charge. Thus, this is given by this mathematical expression:

[tex]q \;\alpha \;L\\\\q=kL\\\\5=k15\\\\k=\frac{5}{15}[/tex]

k = 0.33.

For the -3.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{3.0}{0.33}[/tex]

L = 9.09.

For the -2.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{2.0}{0.33}[/tex]

L = 6.06.

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Answer:

[tex]50.91 \mu C[/tex]

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for [tex]q_{1}[/tex] and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by[tex]q_{1}[/tex] on q. Then we can know the magnitude of the force exerted by [tex]q_{2}[/tex] about q, finally this will allow us to know the magnitude of [tex]q_{2}[/tex]

[tex]q_{1}[/tex] exerts a force on q in +y direction, and [tex]q_{2}[/tex] exerts a force on q in -y direction.

[tex]F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\[/tex]

The net force on q is:

[tex]F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}[/tex]

Rewriting for [tex]q_{2}[/tex]:

[tex]q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C[/tex]

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:

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

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 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) 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:

(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:

56.7°

Explanation:

Imagine a rectangle triangle.

The triangle has 3 sides.

One side is the height of the tower, let's name it A.

Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.

Sides A and B are perpendicular.

The other side is the length of the wire. Let's name it C.

From trigonometry we know that:

cos(a) = B / C

Where a is the angle between B anc C, between the wire and the ground.

Therefore

a = arccos(B/C)

a = arccos(552/1005) = 56.7°

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

Given :

A guy wire 1005 feet long is attached to the top of a tower.

When pulled taut, it touches level ground 552 feet from the base of the tower.

Solution :

We know that,

[tex]\rm cos \theta = \dfrac{Base}{Hypotanuse}[/tex]

Given that,

base = 552 ft

hypotanuse = 1005 ft

Therefore,

[tex]\rm cos\theta = \dfrac{552}{1005}[/tex]

[tex]\rm \theta = cos^-^1 \dfrac{552}{1005}[/tex]

[tex]\theta = 56.7^\circ[/tex]

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

For more information, refer the link given below

https://brainly.com/question/22798381?referrer=searchResults

Answer:371.564 mi

Explanation:

Given

Airplane flies northwest for 250 mi and then travels west 150 mi

That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction

after that it travels 150 mi in -ve x direction

So its position vector is given by

[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]

[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]

so magnitude of displacement is

[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]

|r|=371.564 mi

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].