A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magnitude of the force (in N) on the test charge?
(b) What is the direction of this force (away from or toward the +6 µC charge)?
a. away from the +6 µC charge or
b. toward the +6 µC charge

Answer:

Current through each resistor is 2 A.

Explanation:

Given that,

Voltage of a battery, V = 5 volts

Resistance 1, R = 1.25 ohms

Resistance 2,R' = 1.25 ohms

Both resistors are connected in series. The equivalent resistance is given by :

R" = R + R'

R" = 1.25 + 1.25

R" = 2.5 ohms

The current flowing throughout all resistors is same in series combination of resistors. Current can be calculated using Ohm's law as :

[tex]I=\dfrac{V}{R"}[/tex]

[tex]I=\dfrac{5\ V}{2.5}[/tex]

I = 2 A

So, the current through each resistor is 2 A. Hence, this is the required solution.

Answer:

[tex] v_{max}=14.2\frac{m}{s} [/tex]

Explanation:

Hi!

If the crate is not sliding, its trajectory is the arc with 36.1 m radius. Then the crate  has a centripetal acceleration:

[tex]a_c= \frac{v^2}{r} \\r = radius\\v = tangential \; velocity[/tex]

The centripetal force acting on the crate is the static friction force between crate and truck. The maximum value of this force is:

[tex]F_{max} = \mu N\\\mu = 0.570=static\;friction \;coefficient\\N =normal\; force\\[/tex]

The normal force has a magnitude equal to the weight of the crate:

[tex]N=mg[/tex]

Then the condition for not sliding is:

[tex]F_{centripetal} = M\frac{v^2}{r}<\mu N=\mu Mg\\ v^2<r \mu g = 36.1\;m*0.570*9.8\frac{m}{s^2}= 201.65 \frac{m^2}{s^2}\\ v<14.2\frac{m}{s}[/tex]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Where, [tex]\mu[/tex] = mass per unit length

T = tension

Put the value into the formula

[tex]v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}[/tex]

[tex]v =519.61\ m/s[/tex]

Hence, The speed of transverse waves in this string is 519.61 m/s.

Answer:D-velocity

Explanation:

Given

Constant linear acceleration i.e. acceleration is constant

a=constant

and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]

Thus velocity is changing uniformly to give constant acceleration

While displacement and distance are function of time thus they are not changing uniformly.

During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.

Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.

Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.

Therefore,

During constant linear acceleration, the following quantity changes uniformly: velocity.

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Answer:

75 °F

Explanation:

Air has a specific heat at constant pressure of:

Cpa = 0.24 BTU/(lbm*F)

The specific heat of water is:

Cpw = 1 BTU/(lbm*F)

The first law of thermodynamics:

Q = L + ΔU

The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.

Then:

Q = 0.

We split the heat into the heat transferred by the air and the heat trnasferred by the water:

Qa + Qw = 0

Qa = -Qw

The heat exchanged by the air is

Qa = Ga * Cpa * (tfin - ti)

And the heat exchanged by the water is:

Qw = Gw * Cpw * Δt

Replacing:

Ga * Cpa * (tfin - ti) = -Gw * Cpw * Δt

tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)

tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti

The G terms are mass flows, however we have volume flow of air.

With the gas state equation we calculate the mass:

p * V = m * R * T

m = (p * V) / (R * T)

55 °F = 515 °R

The gas constant for air is R = 53.35 (ft*lb)/(lbm* °R)

14.7 psi = 2117 lb/ft^2

m = (2117 * 5000) / (53.35 * 515) = 385 lbm

The mass flow is that much amount per minute

The mass flow of water is

11200 lbm/h = 186.7 lbm/min

Then:

tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

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

here Final Velocity v=0

a=acceleration due to gravity

[tex]0-u^2=2\left ( -g\right )h[/tex]

[tex]u=\sqrt{2gh}[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]

[tex]u=\sqrt{24.328}[/tex]

u=4.93 m/s

Answer:

Height, h = 300.27 meters

Explanation:

Given that,

Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]

We need to find the height of the manometer required. The pressure at a height is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]

h is the height of the manometer required.

[tex]h=\dfrac{P}{\rho g}[/tex]

[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]

h = 300.27 meters

So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.

Answer:

Explanation:

mass of astronaut, m = 60 kg

mass of space craft, M = 120 kg

t = 1.5 s

Force on space craft = 30 N

(a) According to Newton's third law

Force on spacecraft by the astronaut = Force on astronaut by the space craft

Force on astronaut by the space craft  = 30 N

(b) According to Newton's second law

Force  = mass x acceleration

Let a be the acceleration of the astronaut

30 = 60 x a

a = 0.5 m/s^2

Let A be the acceleration of the spacecraft

30 = 120 x A

a = 0.25 m/s^2

Answer:

Explanation:

So, lets say the runner stars from the position [tex]x_0[/tex]. Lets make this point the origin of a coordinate system in which the vector i points north.

[tex]x_0 = (0,0)[/tex]

Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:

[tex]x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)[/tex].

[tex]x_1 = (20 \ m,0)[/tex].

The, he runs 30 meters south

[tex]x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)[/tex]

[tex]x_2 = (-10 \ m,0)[/tex]

Finally, he runs 40 meter north

[tex]x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)[/tex]

[tex]x_3 = (30 \ m,0)[/tex].

This is our displacement vector. Now, the average speed will be:

[tex]\frac{distance}{time}[/tex].

The distance its the length of the displacement vector,

[tex]d=\sqrt{x^2+y^2}[/tex]

[tex]d=\sqrt{(30 \ m)^2+0^2}[/tex]

[tex]d=30 \ m[/tex]

So, the average speed its:

[tex]\frac{30 \ m }{30 \ s} = 1\frac{m}{s}[/tex].

The average velocity, instead, its:

[tex]\vec{v} = \frac{displacement}{time}[/tex]

[tex]\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}[/tex]

[tex]\vec{v} = (1 \ \frac{m}{s} ,\ 0)[/tex]

This is, 1 m/s north.

Answer:

(a) [tex]E_{ c} = 112.5 \mu J[/tex]

(b) [tex]E'_{ c} = 18 \mu J[/tex]

Solution:

According to the question:

Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]

Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]

(a)The Energy stored in the Capacitor is given by:

[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]

[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]

[tex]E_{ c} = 112.5 \mu J[/tex]

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]

[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]

[tex]E'_{ c} = 18 \mu J[/tex]

(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]

(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]

The given parameters;

The energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]

When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]

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Answer:

The angle is [tex]\theta\approx 14.61[/tex] degrees.

Explanation:

Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:

[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]

If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.

[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].

If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].

[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]

The angle is thus

[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.

Answer:

Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].

Explanation:

Assuming,

Given:

The density of an object is defined as the mass of the object per unit volume.

Therefore,

[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]

Since only 95% of the body of the swimmer is inside the water, therefore,

[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]

According to Archimedes' principle,

[tex]m=m_w\\[/tex]

Using (1),

[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]

Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

[tex]1\ cm=\frac{1}{100}\ m[/tex]

[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]

967500000 years will pass before the radius of the orbit increases by 10%.

Final answer:

The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.

Explanation:

The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.

Answer:

Explanation:

Capacity of a parallel plate capacitor  C = ε₀ A/ d

ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.

C =(  8.85 X10⁻¹² X  27 X 10⁻⁴ ) / 3 X 10⁻³

= 79.65 X 10⁻¹³ F.

potential diff between plate = Charge / capacity

= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³

= 601 V

Electric field = V/d

= 601 / 3 x 10⁻³

= 2 x 10⁵ N/C

Force on proton

= charge x electric field

1.6 x 10⁻¹⁹ x 2 x 10⁵

= 3.2 x 10⁻¹⁴

Acceleration a = force / mass

= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷

= 1.9 x 10¹³ m s⁻²

Distance travelled by proton = 3 x 10⁻³

3 x 10⁻³ = 1/2 a t²

t = [tex]\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }[/tex]

t = 1.77 x 10⁻⁸ s

Answer:

[tex]F_{net}=N\ sin\theta[/tex]

Explanation:

Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.    

The attached free body diagram shows the car on the banked turn. It is clear that,

In vertical direction,

[tex]N\ cos\theta=mg[/tex]

In horizontal direction,

[tex]F_{net}=F_{centripetal}[/tex]

[tex]F_{net}=N\ sin\theta[/tex]

So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).

Hi!

Letw call A to the initially neutral rod, and B to the positively charged. When they are close to each other, the positive charges in B attract the negative charges in A, and repelle the positive ones. If you ground A, negative charges from ground (your body, in this case), flow to A attracted by the positive charges in B, and positive charges in A flow to ground, so finally A results negatively charged

Answer:

Negative charges

Explanation:

The procedure described above is known in physics as charging by electrostatic induction. If we desire to impart negative charges to a hitherto neutral rod, we bring a positively charged rod near it without allowing the two insulated rods to touch each other. If the neutral rod is earthed, negative charges remain on the rod.

Answer:

93750 ft/s²

Explanation:

t = Time taken

u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)

v = Final velocity = 0

s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]

a = Acceleration

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]

The magnitude of acceleration is 93750 ft/s²

Answer:

For 0.24 sec the person was in the air.

Explanation:

Given that,

Height = 1 m

Initial velocity = 3 m/s

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Where, u = initial velocity

s = height

Put the value into the formula

[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]4.9t^2+3t-1=0[/tex]

[tex]t = 0.24\ sec[/tex]

On neglecting the negative value of time

Hence, For 0.24 sec the person was in the air.

Explanation:

Given that,

Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]

Distance between charges, r = 1.81 m

(a) The electrostatic force that one charge exerts on the another is given by :

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

[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]

[tex]F=9.21\times 10^{-8}\ N[/tex]

(b) As both charges are positively charged. So, the force of attraction between them is repulsive.

Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, [tex]g=-9.8\ m/s^2[/tex]

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

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

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

[tex]v^2=2(-9.8)\times 1+(10)^2[/tex]

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

Answer:

8.96 m/s, upward direction

Explanation:

Given that, the initial velocity of the ball is,

[tex]u=10m/s[/tex]

And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,

[tex]a=9.8\frac{m}{s^{2} }[/tex]

And according to question vertical displacement is,

[tex]s=1m[/tex]

Now suppose v be the final velocity of the ball.

Applying third equation of motion,

[tex]v^{2}=u^{2}+2as[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement.

Substitute all the variables.

[tex]v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}[/tex]

Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.

Answer:

The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]

Explanation:

For this particular situation, the angular frequency of the system is given by

[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]

Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.

Plugging the values into the equation:
 vi = d/t
 vi = 1.19 m / 0.20 s
 vi = 5.95 m/s

Therefore, the initial velocity of the ball is approximately 5.95 m/s.

Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.

To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:

t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s

Therefore, the time it takes to reach the peak is approximately 0.61 seconds.

As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:

Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s

Hence, it should take approximately 1.22 seconds before the ball reappears.

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Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance =  3.6 m

v = Image distance

f = Focal length = 35 mm

[tex]h_u[/tex]= Object height = 1.8 m

a) Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm[/tex]

The CCD sensor is 35.34 mm from the lens

b) Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm[/tex]

The person appears 17.67 mm tall on the sensor

Answer:

The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]

Solution:

As per the question:

Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]

Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex]        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]

[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]

[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]

Q = [tex]\pm 5.707\times 10^{13} C[/tex]

Answer:

The life time of the particle is [tex]2.491\times 10^{- 25} s[/tex]

Solution:

As per the question:

Average rest energy of [tex]Z^{0}boson = 91.19 GeV[/tex]

Uncertainty in rest energy, [tex]\Delta E_{r} = 2.5 GeV = 2.5\times 10^{9}\times 1.6\times 140^{- 19} J = 4\times 10^{- 10} J[/tex]

Now,

From the Heisenberg's Uncertainty Principle, we can write:

[tex]\Delta E_{r}\times \Delta T \geq \frac{h}{2\pi}[/tex]

where

T = Life time  of the particle

[tex]\Delta T \geq \frac{h}{2\pi\Delta E_{r}}[/tex]

[tex]\Delta T \geq \frac{6.262\times 10^{- 34}}{2\pi\times 4\times 10^{- 10}}[/tex]

[tex]\Delta T \simeq 2.491\times 10^{- 25} s[/tex]

The lifetime of the Z0 boson is approximately 8.95 x 10^-17 seconds.

The Z0 boson is a particle that mediates the weak nuclear force. Its average rest energy is 91.19 GeV and it has an intrinsic width of 2.5 GeV. The lifetime of a particle can be calculated using the Heisenberg uncertainty principle, which relates the energy uncertainty to the time uncertainty. The relationship is given by the equation ΔE × Δt ≥ ℏ/2, where ℏ is the reduced Planck constant. By rearranging the equation and substituting the values, we can calculate the lifetime of the Z0 boson to be approximately 8.95 x 10^-17 seconds.

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Answer:

Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.

Explanation:

As we know that the potentiometer can act as a variable resistor but resistors have constant value.

In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.

Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.

Answer:

frequency = 19.56 Hz

Explanation:

given data

length L = 7 m

mass m = 7 g

mass M = 2.50 kg

distance d = 4 m

to find out

fundamental frequency

solution

we know here frequency formula is

frequency = [tex]\frac{v}{2d}[/tex]   ...........1

so here d is given = 4

and v = [tex]\sqrt{\frac{T}{\mu} }[/tex]  ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

and μ = [tex]\frac{m}{l}[/tex] =  [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m

so from equation 2

v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]

v = 156.52

and from equation 1

frequency = [tex]\frac{v}{2d}[/tex]

frequency = [tex]\frac{156.52}{2(4)}[/tex]

frequency = 19.56 Hz

Final answer:

The fundamental frequency of the vibrating cord is approximately 14.18 Hz.

Explanation:

To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:

f1 = 1/2L * sqrt(T / μ)

Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.

Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)

Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz

Answer:

Electric field at distance of 50 micrometer due to a proton is 0.576 N/C

Explanation:

We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]

We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]

Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]

Answer:26.96 m,2.34 s

Explanation:

Given

Wrench hit the ground with a speed of 23 m/s

Applying equation of motion

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

Here u=0 because it is dropped from a height of S m

[tex]23^2-0=2\times 9.81\times s[/tex]

[tex]s=\frac{529}{2\times 9.81}=26.96 m[/tex]

Time required by wrench to hit the ground

v=u+at

[tex]23=9.81\times t[/tex]

[tex]t=\frac{23}{9.81}=2.34 s[/tex]

Answer:

(C) a circle starting at time t=0 on the positive x axis

Explanation:

particle's position is

r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^

this is a parametric equation of a circle, because the axis at x and y are the same = R.

for t=0:

r=Ri^

so: circle starting at time t=0 on the positive x axis

On the other hand:

[tex]v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j][/tex]

The value of the magnitude of the acceleration is:

[tex]a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}[/tex]

we can recognise that this represent the centripetal acceleration.