An object with a charge of -2.9 μC and a mass of 1.0×10^−2 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. Find the magnitude of the electric field?
If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.?

Answer:0.249 s

Explanation:

Given

Ball is tossed down with a velocity of 6.8 m/s downward

height from ground=2 m

therefore time to reach ground is

[tex]s=ut+\frac{gt^2}{2}[/tex]

[tex]2=6.8\times t+\frac{9.81\times t^2}{2}[/tex]

[tex]9.81t^2+13.6t-4=0[/tex]

[tex]t=\frac{-13.6\pm \sqrt{13.6^2+4\times 4\times 9.81}}{2\times 9.81}[/tex]

[tex]t=\frac{-13.6+18.49}{19.62}=0.249 s[/tex]

Answer:

Given:

Electric field = 180 N/C

[tex]Force\ on\ proton = 1.6\times10^{-19} C[/tex]

[tex]Force\ on\ proton = 180\times1.6\times10^{-19} =288\times10^{-19} N[/tex]

[tex]Mass\ of\ proton = 1.673\times10^{-27} kg[/tex]

[tex]Acceleration of proton = \frac{force}{mass}[/tex]

[tex]Acceleration\ of\ proton = \frac{288\times10^{-19}}{1.673*10^{-27}} =172\times108 m/s^{2}[/tex]

Let the speed of proton be "x"

x = [tex]\sqrt{Acceleration}[/tex]

[tex]x = \sqrt{(2\times172\times108\times0.125)}=65602.2 m/s[/tex]

Answer:

the velocity of the proton is 65574.38 m/s

Explanation:

given,

uniform electric field = 180 N/C

Distance = 12.5 cm = 0.125 m

charge of proton = 1.6 × 10⁻¹⁹ C

force = E × q

         =180 ×  1.6 × 10⁻¹⁹

        F= 2.88 × 10⁻¹⁷ N

mass of proton = 1.673 × 10⁻²⁷ kg

acceleration =[tex]\dfrac{force}{mass}[/tex]

                     =[tex]\dfrac{2.88 \times 10^{-17}}{1.673\times 10^{-27}}[/tex]

                     =1.72 × 10¹⁰ m/s²

velocity = [tex]\sqrt{2\times 0.125 \times 1.72 \times 10^{10}}[/tex]

             =65574.38 m/s

hence , the velocity of the proton is 65574.38 m/s

Explanation:

Maximum height reached by the ball, s = 52 m

Let u is the initial speed of the ball and v is the final speed of the ball, v = 0 because at maximum height the final speed goes to 0. We need to find u.

(a) The third equation of motion as :

[tex]v^2-u^2=2as[/tex]

Here, a = -g

[tex]0-u^2=-2gs[/tex]

[tex]u^2=2\times 9.8\times 52[/tex]

u = 31.92 m/s

(b) Let t is the time when the ball is in air. It is given by :

[tex]v=u+at[/tex]

[tex]u=gt[/tex]

[tex]t=\dfrac{31.92\ m/s}{9.8\ ms/^2}[/tex]

t = 3.25 seconds

Hence, this is the required solution.                                                                  

Answer:

v = 0.9798*c

Explanation:

E0 = 105 MeV

The mass of a muon is

m = 1.78 * 10^-30 kg

The kinetic energy is:

[tex]Ek = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]

The kinetic energy is 4 times the rest energy.

[tex]4*E0 = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]

[tex]4 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}-1[/tex]

[tex]5 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

[tex]\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{5}[/tex]

[tex]1 - \frac{v^2}{c^2} = \frac{1}{25}[/tex]

v^2 / c^2 = 1 - 1/25

v^2 / c^2 = 24/25

v^2 = 24/25 * c^2

v = 0.9798*c

Answer:[tex]T_L=-154.2^{\circ}[/tex]

Explanation:

Given

COP= 60 % of carnot heat pump

[tex]COP=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]

For heat added directly to be as efficient as via heat pump

[tex]Q_s=W[/tex]

[tex]COP=\frac{Q_s}{W}=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]

[tex]1=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]

[tex]1=\frac{60}{100}\times \frac{24+273}{24+273-T_L}[/tex]

[tex]T_L=118.8 K[/tex]

[tex]T_L=-154.2^{\circ}[/tex]

Answer:

1.57 kW

Explanation:

The rate of heat loss is given by:

q = Gm * Cp * (tfin - ti)

Where

q: rate of heat loss

Gm: mass flow

Cp: specific heat at constant pressure

The Cp of air is:

Cp = 1 kJ/(kg*K)

The mass flow is the volumetric flow divided by the specific volume

Gm = Gv / v

The volumetric flow is the air speed multiplied by the cruss section of the duct.

Gv = s * h * w (I name speed s because I have already used v)

The specific volume is obtained from the gas state equation:

p * v = R * T

60 C is 333 K

The gas constant for air is 287 J/(kg*K)

Then:

v = (R * T)/p

v = (287 * 333) / 100000 = 0.955 m^3/kg

Then, the mass flow is

Gm = s * h * w / v

And rthe heat loss is of:

q = s * h * w * Cp * (tfin - ti) / v

q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)

Answer:

Accuracy

Explanation:

Percent error is the ratio of the difference of the measured and actual value to  the actual value multiplied by 100.

It gives the percent deviation of the value obtained from the actual value.

Accuracy is the measure of how close the readings are to the actual value or set standard and can be improved by increase the no. of readings in an experiment.

Precision is the measure of the closeness of the obtained values to one another.

Thus accuracy of the reading can be sensed by the percent error.

Answer:

The height of mountain in meter will be 106.2732 m

Explanation:

We have given height of mountain = 348 ft,8 in

We know that 1 feet = 0.3048 meter

So 348 feet [tex]=348\times 0.3048=106.07meter[/tex]

And we know that 1 inch = 0.0254 meter

So 8 inch [tex]8\times 0.0254=0.2032m[/tex]

So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m

The height of mountain in meter will be 106.2732 m

Answer:d-6.7 km

Explanation:

Given

Bicyclist pedals at a speed of 5 km/h

so his speed in meter per second

[tex]5\times \frac{5}{18}=\frac{25}{18} m/s[/tex]

In 80 minutes he would travel

Distance traveled[tex]=\frac{25}{18}\times 80\times 60=6666.667 m\approx 6.67 km[/tex]

1) 9.18 s

In the first part of the motion, the rocket accelerates at a rate of

[tex]a_1=13.5 m/s^2[/tex]

For a time period of

[tex]t_1=3.50 s[/tex]

So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:

[tex]v_1=u+a_1t_1[/tex]

where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,

[tex]v_1=(13.5)(3.50)=47.3 m/s[/tex]

In the second part of the motion, the rocket decelerates with a constant acceleration of

[tex]a_2 = -5.15 m/s^2[/tex]

Until it comes to a stop, to reach a final velocity of

[tex]v_2 = 0[/tex]

So we can use again the same equation

[tex]v_2 = v_1 + a_2 t_2[/tex]

where [tex]v_1 = 47.3 m/s[/tex]. Solving for t2, we find after how much time the rocket comes to a stop:

[tex]t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s[/tex]

2) 299.9 m

We have to calculate the distance travelled by the rocket in each part of the motion.

The distance travelled in the first part is given by:

[tex]d_1 = ut_1 + \frac{1}{2}a_1 t_1^2[/tex]

Using the numbers found in part a),

[tex]d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m[/tex]

The distance travelled in the second part of the motion is

[tex]d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2[/tex]

Using the numbers found in part a),

[tex]d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m[/tex]

So, the total distance travelled by the rocket is

d = 82.7 m + 217.2 m = 299.9 m

Final answer:

The speed of the block is 4.08 m/s, the magnitude of the block’s acceleration is 25.41 m/s^2, and the direction of the block’s acceleration is toward the wall.

Explanation:

(a) To find the speed of the block, we can use the principle of conservation of mechanical energy. The potential energy stored in the spring when it is compressed is converted into the kinetic energy of the block when it is released. The potential energy stored in the spring is given by:

PE = 0.5 * k * x^2

where k is the force constant of the spring and x is the compression of the spring. Plugging in the values, we get:

PE = 0.5 * 130.0 N/m * 0.80 m * 0.80 m = 41.60 J

The kinetic energy of the block when it is released is given by:

KE = 0.5 * m * v^2

where m is the mass of the block and v is its speed. Equating the potential and kinetic energies, we have:

PE = KE

41.60 J = 0.5 * 3.00 kg * v^2

Solving for v, we get:

v = √(41.60 J / (0.5 * 3.00 kg)) = 4.08 m/s

(b) The magnitude of the block's acceleration can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force applied to the block minus the force of friction. The force applied to the block is given by F = 88.0 N. The force of friction can be calculated using the equation:

f = μk * m * g

where μk is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values, we get:

f = 0.400 * 3.00 kg * 9.8 m/s^2 = 11.76 N

The net force is therefore:

net force = F - f = 88.0 N - 11.76 N = 76.24 N

Using Newton's second law, we have:

76.24 N = 3.00 kg * a

Solving for a, we get:

a = 76.24 N / 3.00 kg = 25.41 m/s^2

(c) The direction of the block's acceleration can be determined by considering the net force acting on the block. In this case, the applied force and the force of friction are in opposite directions, resulting in a net force in the direction of the applied force. Therefore, the direction of the block's acceleration is toward the wall.

Answer:

The energy transmitted per second down the wire is 0.761 watt.

Explanation:

Given that,

Length = 2.5 m

Amplitude = 3.0 mm

Density = 20 g/m

Frequency = 35 Hz

We need to calculate the wavelength

Using formula of wavelength

[tex]L = \dfrac{\lambda}{2}[/tex]

[tex]\lambda=2L[/tex]

Put the value into the formula

[tex]\lambda=2\times2.5[/tex]

[tex]\lambda=5\ m[/tex]

We need to calculate the speed

Using formula of speed

[tex]v = f\lambda[/tex]

Put the value into the formula

[tex]v =35\times5[/tex]

[tex]v =175\ m/s[/tex]

We need to calculate the energy is transmitted per second down the wire

Using formula of the energy is transmitted per second

[tex]P=\dfrac{1}{2}\mu A^2\omega^2\times v[/tex]

[tex]P=\dfrac{1}{2}\mu\times A^2\times(2\pi f)^2\times v[/tex]

Put the value into the formula

[tex]P=\dfrac{1}{2}\times20\times10^{-3}\times(3.0\times10^{-3})^2\times4\times\pi^2\times(35)^2\times175[/tex]

[tex]P=0.761\ watt[/tex]

Hence, The energy transmitted per second down the wire is 0.761 watt.

Answer:

a) The acceleration took 0.0076s

b) The aceleration was of 7202.4 m/s^2

Explanation:

We need to use the formulas for acceleration movement in straight line that are:

(1) [tex]a = \frac{V}{t}[/tex]    and  (2)[tex]x=x_{0} +V_{0}t + \frac{1}{2} at^2[/tex]

Where

a = acceleration

V = Velocity reached

Vo = Initial velocity

t = time

x = distance

xo = initial distance.

We have the following information:

a = We want to find      V = 55.0 m/s      

Vo = 0m/s because it starts from rest       t = we want to find      

x = 0.210 m         xo= 0 m we beging in the point zero.

We have to variables in two equations, so we are going to replace in the second equation (2) the aceleration of the first one(1):

[tex]x=x_{0} +V_{0}t + \frac{1}{2} ( \frac{V}{t})t^2[/tex] We can cancel time because it is mutiplying and dividing the same factor so we have

[tex]x=x_{0} +V_{0}t + \frac{1}{2} Vt[/tex]    

In this equation we just have one variable that we don't know that is time, so first we are going to replace the values and after that clear time.

[tex]0.210=0 +0*t + \frac{1}{2} 55t[/tex]

[tex]0.210=27.5t[/tex]

[tex]\frac{0.21}{27.5} = t\\[/tex]

t = 0.0076s

a) The acceleration took 0.0076s

Now we replace in the (1) equation the values of time and velocity

[tex]a = \frac{V}{t}[/tex]

[tex]a = \frac{55}{0.0076}[/tex]

a = 7202.4 m/s^2

b) The aceleration was of 7202.4 m/s^2

Answer:

[tex]Q = -1.43\times 10^[-5} coulomb[/tex]

Explanation:

Given data:

particle mass =  0.923 g

particle charge is 4.52 micro C

speed of particle 45.7 m/s

In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve

[tex]-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}[/tex]

solving for Q WE GET

[tex]Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}[/tex]

[tex]Q = -mv^2\times r \frac{4\pi \epsilon}{q}[/tex]

[tex]Q = - \frac{0.923\times 10^{-3} \times 45.7^2\times (22.6\times 10^{-2})} {4.52\times 10^{-6} \times 9\times 10^9}[/tex]

where[tex] \frac{1}{4\pi \epsilon} = 9\times 10^9[/tex]

[tex]Q = -1.43\times 10^[-5} coulomb[/tex]

Final answer:

To find the charge Q for circular motion, equate centripetal force m * v^2 / r with Coulomb's force k * |Q * q| / r^2, and solve for Q. Use m = 0.923 g, v = 45.7 m/s, q = 4.52 µC, and convert units accordingly.

Explanation:

To determine the value of charge Q that will allow the moving particle to execute circular motion, we use the concept of centripetal force. Centripetal force is the net force required to keep an object moving in a circle at a constant speed and is directed towards the center of the circle. For a charged particle moving in a circular path due to an electric force, the centripetal force is provided by the electric force between the charges.

The centripetal force (Fc) is equal to the mass (m) of the particle times the square of its speed (v) divided by the radius (r) of the circle:
Fc = m * v2 / r.
The electric force (Fe) acting on the particle is given by Coulomb's law:
Fe = k * |Q * q| / r2,
where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge at the origin, q is the charge of the moving particle, and r is the separation between the charges.

Setting the centripetal force equal to the electric force yields:
m * v2 / r = k * |Q * q| / r2,
Solving for Q gives us:
Q = m * v2 / (k * q).

Plugging in the values:
Q = (0.923 g * 45.7 m/s2) / (8.99 x 109 Nm2/C2 * 4.52 µC)
Remembering to convert grams to kilograms and microcoulombs to coulombs, the final calculation will yield the required charge Q for circular motion.

Q = 1.03 mC

Answer:

a) 1.2*10^-4 m/s

b) 375 m/s

Explanation:

I assume the large asteroid doesn't move.

The smaller asteroid is affected by an acceleration determined by the universal gravitation law:

a = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s^2))

M: mass of the large asteroid (33900 kg)

d: distance between them (146 m)

Then:

a = 6.67*10^-11 * 33900 / 146^2 = 10^-10 m/s^2

I assume the asteroid in a circular orbit, in this case the centripetal acceleration is:

a = v^2/r

Rearranging:

v^2 = a * r

[tex]v = \sqrt{a * r}[/tex]

v = \sqrt{10^-10 * 146} = 1.2*10^-4 m/s

If the asteroids have electric charges of 1.18 C and -1.18 C there will be an electric force of:

F = 1/(4π*e0)*(q1*q2)/d^2

Where e0 is the electrical constant (8.85*10^-12 F/m)

F = 1/(4π*8.85*10^-12) (-1.18*1.18)/ 146^2 = -587 kN

On an asteroid witha mass of 610 kg this force causes an acceleration of:

F = m * a

a = F / m

a = 587000 / 610 = 962 m/s^2

With the electric acceleration, the gravitational one is negligible.

The speed is then:

v = \sqrt{962 * 146} = 375 m/s

The number of electrons lost by the by the honeybee in acquiring the charge of +20 pC is;

n = 1.25 × 10^(8) electrons

We are given;

Charge of honeybee; Q = 20 pC = 20 × 10^(-12) C

Now, formula for number of electrons is;

n = Q/e

Where;

e is charge on electron = 1.6 × 10^(-19) C

Thus;

n = (20 × 10^(-12))/(1.6 × 10^(-19))

n = 1.25 × 10^(8) electrons

Read more at; https://brainly.com/question/14653647

Answer:

option A is correct

25% free lead and 75% lead oxide

Explanation:

we have given free lead and lead oxide %

so here we know lead oxide material usually composed in the range of for free lead and lead oxide as

lead oxide material contain free lead range is 22 % to 38 %

so here we have only option 1 which contain in free lead in the range of 22 % to 38 % i.e 25 % free lead

so    

option A is correct

25% free lead and 75% lead oxide

Answer:

The velocity waves before rain is 10 m/s

The velocity of wave after the rope soaked up 5 kg more is 8.944 m/s

Solution:

As per the question:

Length of the rope, l = 100 m

Mass of the rope, m = 20 kg

Force due to tension in the rope, [tex]T_{r} = 20 N[/tex]

Frequency of vibration in the rope, f = 10 Hz

Extra mass of the rope after being soaked in rain water, m' = 5 kg

Now,

In a rope, the wave velocity is given by:

[tex]v_{w} = \sqrt{\frac{T_{r}}{M_{d}}}[/tex]         (1)

where

[tex]M_{d}[/tex] = mass density

Mass density before soaking, [tex]M_{d} = \frac{m}{l} = \frac{20}{100} = 0.20[/tex]

Mass density after being soaked, [tex]M_{d} = \frac{m + m'}{l} = \frac{25}{100} = 0.25[/tex]

Initially, the velocity is given by using eqn (1):

[tex]v_{w} = \sqrt{\frac{20}{0.20}} = 10 m/s[/tex]

The velocity after being soaked in rain:

[tex]v_{w} = \sqrt{\frac{20}{0.25}} = 8.944 m/s[/tex]

To find out how high the passenger is when the camera reaches her, we use kinematic equations, taking into account the initial speed of the camera, the constant rise speed of the passenger, and gravity's acceleration. The solution requires equating the displacements of both camera and passenger to solve for time and therefore the height.

A hot-air balloon is rising at a constant rate of 2.0m/s when a passenger's camera is tossed straight upward with an initial speed of 12m/s from a position 2.5m below her. To determine how high the passenger is when the camera reaches her, we can apply kinematic equations of motion, incorporating the constant acceleration due to gravity (approximately 9.81m/s² downwards).

For the camera: Its initial upward velocity is 12 m/s, and it is subject to gravity's acceleration. For the passenger: Rising at a constant 2.0 m/s, not accelerating since the rate is constant. Since the initial distance between them is 2.5 m, we need to calculate when the camera, starting from a lower point but moving faster, reaches the vertically moving passenger.

Using the formula s = ut + 0.5at² for both camera and passenger, where s is the displacement, u is initial velocity, a is acceleration, and t is time, we can set the equations equal to solve for t, then determine the height by applying it to the passenger's motion equation.

Due to the mathematical complexity and potential for variability in solving these equations, the exact numerical solution isn't presented here. However, the approach involves determining the time it takes for the camera to reach the same height as the passenger and using that to find her height at that moment.

Answer:

average velocity = 6.25m/sec

Explanation:

given data:

for unopened

height = 625 m

time  = 15 sec

for opened

height = 356 m

time =  142 sec

Unopened:

[tex]V1 = \frac{625\ m}{15\ sec} = 41.67m/sec[/tex]

Opened:

[tex]V2 = \frac{356\ m}{142\ sec} = 2.51m/sec[/tex]

we know that

Total Average Velocity[tex] = \frac{Total\ distance}{Total\ time}[/tex]

average velocity[tex] = \frac{981\ m}{157\ sec}[/tex]

average velocity = 6.25m/sec

downward direction.

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

[tex]P_1, V_1, T_1 & P_2, V_2, T_2 [/tex]

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

[tex]-Q=W+U_1----1[/tex]

[tex]Q=-W+U_2-----2[/tex]

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

[tex]U_1 & U_2[/tex] change in internal Energy of gas

Thus from 1 & 2 we can say that

[tex]U_1+U_2=0[/tex]

[tex]n_1c_v(T-T_1)+n_2c_v(T-T_2)=0[/tex]

[tex]T(n_1+n_2)=n_1T_1+n_2T_2[/tex]

[tex]T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}[/tex]

where [tex]n_1=\frac{P_1V_1}{RT_1}[/tex]

[tex]n_2=\frac{P_2V_2}{RT_2}[/tex]

[tex]T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}[/tex]

[tex]T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}[/tex]

and [tex]P=\frac{P_1V_1+P_2V_2}{V_1+V_2}[/tex]

Answer:

[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

[tex]\dfrac{1}{2}mV^2=mg(X-X_o)[/tex]

So

[tex]V=\sqrt{2g(X-X_o)}[/tex]

We know that

[tex]\dfrac{dX}{dt}=V[/tex]

[tex]\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}[/tex]

[tex]\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX[/tex]

At t= 0 ,X=Xo

So we can say that

[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]

So the length of cable after time t

[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]

Answer:

Xylene boils at 138.5 °C, 740.97 R and 411.65 K

Explanation:

[tex]T_{\°C}=(T_{\°F}-32)\cdot \frac{5}{9}[/tex]

We know that temperature is 281.3 °F so in °C is:

[tex]\°C=(281.3-32)\cdot \frac{5}{9}= 138.5 \°C[/tex]

[tex]T_{R}=T_{\°F}+459.67\\T_{R}=281.3\°F+459.67=740.97 R[/tex]

[tex]T_{K}=(T_{\°F}+459.67)\cdot \frac{5}{9} \\T_{K}=(281.3 \°F+459.67)\cdot \frac{5}{9} \\T_{K}=411.65K[/tex]

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

[tex]v = C\dfrac{\pm \dlta f}{2 f_0}[/tex]

[tex]\delta f = \pm 2 f_0 (\dfrac{V}{C})[/tex]

[tex]\delta f = \pm 2\times 7 (\dfrac{4}{1500})[/tex]

           [tex]=\pm 0.0373 MHz[/tex]

           = 37.3 kHz

|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

Answer:

Explanation:

distance between slits d = 1 x 10⁻³ m

Screen distance D = 1.2 m

Wave length of light   = λ

Distance of n th bright fringe fro centre

= n λ D / d where n is order of bright fringe . Here n = 4

Given

3.1 x 10⁻³ = (4 x λ x 1.2) / 1 x 10⁻³

λ = 3.1 x 10⁻⁶ / 4.8

= .6458 x 10⁻⁶

6458 x 10⁻¹⁰m

λ= 6458 A.

The distance will reduce 1.33 times

New distance = 3.1 /1.33

= 2.33 mm.

Answer:

a) 1.29*10^6 N/C

b) 0.703 *10^6 m/s

Explanation:

This is a parallel plates capacitor. In a parallel plates capacitor the electric field depends on the charge of the disks, its area and the vacuum permisivity (Assuming there is no dielectric) and can be found using the expression:

[tex]E = \frac{Q}{A*e_0} =\frac{11*10^{-9}C}{(\frac{1}{4}\pi*(0.035m)^2)*8.85*10^{-12}C^2/Nm^2} = 1.29 *10^6 N/C[/tex]

For the second part, we use conservation of energy. The change in kinetic energy must be equal to the change in potential energy. The potential energy is given by:

[tex]PE = V*q[/tex]

V is the electric potential or voltage, q is the charge of the proton. The electric potential is equal to:

[tex]V = -E*d[/tex]

Where d is the distance to the positive disk. Then:

[tex]\frac{1}{2}mv_1^2 +V_1q = \frac{1}{2}mv_2^2 +V_2q\\\frac{1}{2}m(v_1^2 - v_2^2)=(V_2-V_1)q = (r_1-r_2)Eq|r_2 = 0m, v_2=0m/s\\v_1 = \sqrt{2\frac{(0.002m)*1.29*10^6 N/C*1.6*10^{-19}C}{1.67*10^{-27}kg}}= 0.703 *10^6 m/s[/tex]

Answer:

The maximum speed of car will be 20.5m/sec

Explanation:

We have given mass of car = 1100 kg

Radius of curve = 82.3 m

Static friction [tex]\mu _s=0.521[/tex]

We have to find the maximum speed of car

We know that at maximum speed centripetal force will be equal to frictional force [tex]m\frac{v^2}{r}=\mu _srg[/tex]

[tex]v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec[/tex]

So the maximum speed of car will be 20.5m/sec

Answer:20.51 m/s

Explanation:

Given

Mass of car(m)=1100 kg

radius of curve =82.3 m

coefficient of static friction([tex]\mu [/tex])=0.521

here centripetal force is provided by Friction Force

[tex]F_c(centripetal\ force)=\frac{mv^2}{r}[/tex]

Friction Force[tex]=\mu N[/tex]

where N=Normal reaction

[tex]\frac{mv^2}{r}=\mu N[/tex]

[tex]\frac{1100\times v^2}{82.3}=0.521\times 1100\times 9.81[/tex]

[tex]v^2=0.521\times 9.81\times 82.3[/tex]

[tex]v=\sqrt{420.63}=20.51 m/s [/tex]

Answer:

The chemical energy of the battery was reduced in 10800J

Explanation:

The first thing to take into account is that the stored energy in a battery is in Watts per second or Joules ([tex]W\cdot s=J[/tex]). It means that the battery provides a power for a certain time.

The idea is to know how much [tex]W\cdot s[/tex] has been consumed by the circuit.

The first step is to know the power that is consumed by the circuit. It is [tex]P=V\cdot I[/tex]. The problem says that the circuit consumes a current of 5.0A with a voltage of 6.0V. It means that the power consumed is:

[tex]P=V\cdot I=(6.0V)\cdot (5.0A)=30W[/tex]

The previous value (30W) is the power that the circuit consumes.

Now, you must find the total amount of power that is consumed by the circuit in 6.0 minutes. You just have to multiply the power that the circuit consumed by the time it worked, it means, 6.0 minutes.

[tex]energy=P\cdot t=(30W)\cdot (6.0min)=180W\cdot min[/tex]

You must convert the minutes unit to seconds. Remember that 1 minute has 60 seconds.

[tex]energy=P\cdot t=(30W)\cdot (6.0min) \cdot \frac{60s}{1min}=10800W\cdot s=10800J[/tex]

Thus, the chemical energy of the battery was reduced in 10800J

Final answer:

To calculate the muzzle velocity, we use the equation for constant acceleration (v = u + at) with an initial velocity (u) of 0, the given acceleration (a) of 5.70 x 10^5 m/s^2, and the time (t) of 9.60 x 10^-4 s to find the final velocity (v), which is approximately 547 m/s.

Explanation:

The question concerns calculating the muzzle velocity of a bullet as it's accelerated from the firing chamber to the end of the barrel of a gun.

Using the formula v = u + at, where v is the final velocity, u is the initial velocity (which is zero before the gun is fired), a is the acceleration, and t is the time for which the acceleration is applied, we can find the muzzle velocity. The bullet experiences an acceleration (a) of 5.70 x 105 m/s2 for a time (t) of 9.60 x 10−4 seconds.

Plugging the values into the formula, we get:

v = 0 + (5.70 x 105 m/s2)(9.60 x 10−4 s)

v = 5.47 x 102 m/s

Therefore, the final muzzle velocity of the bullet as it leaves the barrel is approximately 547 m/s.

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

[tex]Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi[/tex]

[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi[/tex]

[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi[/tex]

[tex]Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi[/tex]

[tex]Q = \frac{\pi^2 r^4}{2}}[/tex]

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)