A 14-Ω coffee maker and a 14-Ω frying pan are connected in series across a 120-V source of voltage. A 20-Ω bread maker is also connected across the 120-V source and is in parallel with the series combination. Find the total current supplied by the source of voltage.

The distance traveled by a particle with velocity function

over the first t seconds can be obtained by integrating that function from 0 to t using integration by parts.

The distance traveled by a particle can essentially be obtained by integrating the velocity function over a given timeframe. In this case, the velocity function is given by

To calculate the distance traveled over the first t seconds, we need to integrate this function from 0 to t. Therefore, the integral ∫v(t)dt from 0 to t will provide the required solution. Use the method of integration by parts, choosing u=t^2 and

then complete the integration process following the standard procedure.

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Answer:

Torque, [tex]\tau=0.1669\ N-m[/tex]

Explanation:

It is given that,

Radius of the circular loop, r = 0.7 cm = 0.007 m

Number of turns, N = 520

Current in the loop, I = 3.9 A

The axis of the loop makes an angle of 57 degrees with a magnetic field.

Magnetic field, B = 0.982 T

We need to find the magnitude of torque on the loop. It is given by :

[tex]\tau=\mu\times B[/tex]

[tex]\tau=NIABsin(90-57)[/tex]

[tex]\tau=520\times 3.9\ A\times \pi (0.007\ m)^2\times 0.982\ T\ cos(57)[/tex]

[tex]\tau=0.1669\ N-m[/tex]

[tex]\tau=0.167\ N-m[/tex]

So, the magnitude of torque is 0.1669 N-m. Hence, this is the required solution.

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        [tex]\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}[/tex]

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

[tex]a=\frac{11.7-26.5}{5.2}=-2.85m/s^2[/tex]

Angular acceleration

        [tex]\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2[/tex]

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

Answer:

The height is 1,225 meters

Explanation:

DistanceX= speedX × time ⇒ time= (5 meters) ÷ (10 meters/second) = 0,5 seconds

DistanceY= high= (1/2) × g × (time^2) = (1/2) × 9,8 (meters/(second^2)) × 0,25 (second^2) = 1,225 meters

Answer:

Final angular speed equals 3 revolutions per second

Explanation:

We shall use conservation of angular momentum principle to solve this problem since the angular momentum of the system is conserved

[tex]L_{disk}=I_{disk}\omega \\\\L_{disk}=\frac{1}{2}mr^{2}\\\therefore L_{disk}=\frac{1}{2}mr^{2}\times10rad/sec[/tex]

After the disc and the dropped rod form a single assembly we have the final angular momentum of the system as follows

[tex]L_{final}=I_{disk+rod}\times \omega_{f} \\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}mL_{rod}^{2}\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}m\times (2r_{disc})^{2}\\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{3}mr_{disc}^{2}\\\\L_{final}=\frac{5mr_{disc}^{2}}{6}\times \omega _{f}\\\\[/tex]

Equating initial and final angular momentum we have

[tex]\frac{5mr_{disc}^{2}}{6}\times \omega _{f}=\frac{1}{2}m_{disc}\times r_{disc}^{2}\times 10\pi rad/sec[/tex]

Solving for [tex]\omega_{f}[/tex] we get

[tex]\omega_{f}=6\pi rad/sec[/tex]

Thus no of revolutions in 1 second are 6π/2π

No of revolutions are 3 revolutions per second

Explanation:

It is given that,

Mass of the metal wire, m = 500 g = 0.5 kg

Tension in the wire, T = 80 N

Length of wire, l = 50 cm = 0.5 m

(a) The speed of the transverse wave is given by :

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

M is the mass per unit length or M = m/l

[tex]v=\sqrt{\dfrac{T.l}{m}}[/tex]

[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{0.5\ kg}}[/tex]

v = 8.94 m/s

(b) If the wire is cut in half, so l = l/2

[tex]v=\sqrt{\dfrac{T.l}{2m}}[/tex]

[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{2\times 0.5\ kg}}[/tex]

v = 6.32 m/s

Hence, this is the required solution.

The wavelengths in the system's emission spectrum, in ascending order, are [tex]\(146 \, \text{nm}\) and \(249 \, \text{nm}\).[/tex]

To find the wavelengths associated with the allowed energies of the quantum system, we can use the formula for the energy of a photon:

[tex]\[ E = \frac{hc}{\lambda} \][/tex]

where:

-[tex]\( E \)[/tex] is the energy of the photon,

- [tex]\( h \)[/tex] is Planck's constant[tex](\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),[/tex]

- [tex]\( c \)[/tex] is the speed of light [tex](\( 3.00 \times 10^8 \, \text{m/s} \)),[/tex]

- [tex]\( \lambda \)[/tex] is the wavelength of the photon.

Given the energies [tex]\(0.0 \, \text{eV}\), \(5.0 \, \text{eV}\), and \(8.5 \, \text{eV}\)[/tex], we need to convert these energies to joules, since the units in the formula for energy are in joules.

1.[tex]\(0.0 \, \text{eV}\) corresponds to \(0.0 \, \text{J}\),[/tex]

2. [tex]\(5.0 \, \text{eV}\) corresponds to \(5.0 \times 1.602 \times 10^{-19} \, \text{J}\),[/tex]

3. [tex]\(8.5 \, \text{eV}\) corresponds to \(8.5 \times 1.602 \times 10^{-19} \, \text{J}\).[/tex]

Now, we can use these energies to calculate the wavelengths of the photons:

1. For [tex]\(0.0 \, \text{J}\):[/tex]

[tex]\[ \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{0.0 \, \text{J}}} \][/tex]

2. For [tex]\(5.0 \times 1.602 \times 10^{-19} \, \text{J}\)[/tex]:

[tex]\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{5.0 \times 1.602 \times 10^{-19} \, \text{J}} \]\[ \lambda \approx \frac{1.995 \times 10^{-25}}{5.0 \times 1.602} \, \text{m} \]\[ \lambda \approx 2.49 \times 10^{-8} \, \text{m} \][/tex]

3. For [tex]\(8.5 \times 1.602 \times 10^{-19} \, \text{J}\):[/tex]

[tex]\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{8.5 \times 1.602 \times 10^{-19} \, \text{J}} \]\[ \lambda \approx \frac{1.995 \times 10^{-25}}{8.5 \times 1.602} \, \text{m} \]\[ \lambda \approx 1.46 \times 10^{-8} \, \text{m} \][/tex]

Now, let's convert these wavelengths to nanometers:

[tex]\( 2.49 \times 10^{-8} \, \text{m} = 249 \, \text{nm} \),[/tex]

[tex]. \( 1.46 \times 10^{-8} \, \text{m} = 146 \, \text{nm} \)[/tex]

So, the wavelengths in the system's emission spectrum, in ascending order, are [tex]\(146 \, \text{nm}\) and \(249 \, \text{nm}\).[/tex]

Answer:

t = 4.35 s

Explanation:

Since the balloon is moving upwards while the sand bag is dropped from it

so here the velocity of sand bag is same as the velocity of balloon

so here we can use kinematics to find the time it will take to reach the ground

[tex]\Delta y = v_y t + \frac{1}{2} gt^2[/tex]

here we know that since sand bag is dropped down so we have

[tex]\Delta y = -85.8 m[/tex]

initial upward speed is

[tex]v_y = 1.55 m/s[/tex]

also we know that gravity is downwards so we have

[tex]a = - 9.8 m/s^2[/tex]

so here we have

[tex]-85.8 = 1.55 t - \frac{1}{2}(9.8) t^2[/tex]

[tex]4.9 t^2 - 1.55 t - 85.8 = 0[/tex]

[tex]t = 4.35 s[/tex]

Answer:

3.08 Nm

Explanation:

N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree

The angle made with the normal of the coil, theta = 90 - 30 = 60 degree

Torque = N I A B Sin Theta

Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60

Torque = 3.08 Nm

Answer:

[tex]T_c=-210.15^{\circ}C[/tex]

Explanation:

In this question we need to convert the temperature in kelvin to degree Celsius. The conversion from kelvin scale to Celsius scale is as follows :

[tex]T_k=T_c+273.15[/tex]

Here,

[tex]T_k=63\ K[/tex]

[tex]T_k-273.15=T_c[/tex]

[tex]63-273.15=T_c[/tex]

[tex]T_c=-210.15^{\circ}C[/tex]

Here, negative sign shows that the heat is released. So, the temperature at 63 K is equivalent to 210.15 °C. Hence, this is the required solution.

Explanation:

Charge of electron in He, [tex]q_e=1.6\times 10^{-19}\ kg[/tex]

Charge of proton in He, [tex]q_p=1.6\times 10^{-19}\ kg[/tex]

Distance between them, [tex]d=2.7\times 10^{-10}\ m[/tex]

We need to find the electric force between them. It is given by :

[tex]F=k\dfrac{q_eq_p}{d^2}[/tex]

[tex]F=-9\times 10^9\times \dfrac{(1.6\times 10^{-19}\ C)^2}{(2.7\times 10^{-10}\ m)^2}[/tex]

[tex]F=-3.16\times 10^{-9}\ N[/tex]

Since, there are two protons so, the force become double i.e.

[tex]F=2\times 3.16\times 10^{-9}\ N[/tex]

[tex]F=6.32\times 10^{-9}\ N[/tex]

So, the correct option is (c). Hence, this is the required solution.

Answer:

A) 50.0 g Al

Explanation:

We can calculate the temperature change of each substance by using the equation:

[tex]\Delta T=\frac{Q}{mC_s}[/tex]

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

[tex]C_s[/tex] is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

[tex]\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C[/tex]

B) m = 50.0 g, Cs = 0.385 J/g°C

[tex]\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C[/tex]

C) m = 25.0 g, Cs = 0.79 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C[/tex]

D) m = 25.0 g, Cs = 0.128 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C[/tex]

E) m = 25.0 g, Cs = 0.235 J/g°C

[tex]\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C[/tex]

As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.

The substance that would show the smallest temperature change upon gaining 200.0 J of heat is Au (Gold), as calculated using the formula for calculating heat (Q = mcΔT) and rearranging for ΔT, then substituting the given values.

The substance that would show the smallest temperature change upon gaining 200.0 J of heat can be determined using the formula used to calculate heat (Q), which is Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change. We want to find the smallest temperature change, so we rearrange the equation to solve for ΔT, which gives us ΔT = Q/(mc). By substituting the given values for each substance into this equation, we find that the smallest temperature change is for Au (Gold).

For Au: ΔT = 200.0J / (25.0g x 0.128 J/g°C) = 62.5°C. All other substances have a smaller temperature change when they absorb 200.0J of heat, due to their higher specific heat capacity.

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Answer:

[tex]q = 0.107 \mu C[/tex]

Explanation:

As we know that net force is given by

[tex]F = ma[/tex]

here we have

m = 1.00 g = 0.001 kg

also we know that acceleration is given as

[tex]a = 256 m/s^2[/tex]

now force is given as

[tex]F = 0.001(256) = 0.256 N[/tex]

now by the formula of force we know that

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

[tex]0.256 = \frac{(9\times 10^9)q^2}{(0.02)^2}[/tex]

now for solving charge we have

[tex]q = 0.107 \mu C[/tex]

The magnitude of the angular momentum of the paperclip attached to the spinning vinyl record is approximately 0.0033 kg m²/s. This is calculated using the formulas for moment of inertia and angular momentum.

To calculate the angular momentum of the paperclip, we first need to know the moment of inertia (I) of the paperclip. The moment of inertia can be calculated using the formula I = mR² where 'm' is the mass of the paperclip (converted into kg - 0.018 kg) and 'R' is the radius of the record player (converted into m - 0.24 m).

So, I = 0.018 kg * (0.24 m)² = 0.0010368 kg m².

Next, we use the formula for angular momentum (L), which is L = Iω, where ω is the angular velocity. Given ω = 3.2 rad/s, we plug these values into our formula:

L = 0.0010368 kg m² * 3.2 rad/s = t0.00331776 kg m²/s.

Thus, rounding to the nearest ten-thousandth, the magnitude of the angular momentum of the paperclip is 0.0033 kg m²/s.

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The magnitude of the angular momentum of the paper clip is approximately [tex]\( 0.0033 \text{ kg m}^2/\text{s} \)[/tex].

The magnitude of the angular momentum of the paper clip is given by the formula [tex]\( L = I\omega \)[/tex], where I is the moment of inertia of the paper clip and [tex]\( \omega \)[/tex] is the angular velocity of the record.

Given:

- Mass of the paper clip, [tex]\( m = 18 \) g \( = 0.018 \)[/tex] kg (after converting grams to kilograms)

- Diameter of the record, [tex]\( d = 48 \) cm \( = 0.48 \)[/tex] m (after converting centimeters to meters)

- Radius of the record, [tex]\( r = \frac{d}{2} = \frac{0.48}{2} = 0.24 \)[/tex]m

- Angular velocity, [tex]\( \omega = 3.2 \)[/tex] rad/s

Now, we calculate the moment of inertia I:

[tex]\[ I = mr^2 = 0.018 \times (0.24)^2 \] \[ I = 0.018 \times 0.0576 \] \[ I = 0.0010368 \text{ kg m}^2 \][/tex]

Next, we calculate the angular momentum L:

[tex]\[ L = I\omega \] \[ L = 0.0010368 \times 3.2 \] \[ L = 0.00331776 \text{ kg m}^2/\text{s} \][/tex]

Rounding to the nearest ten-thousandth, we get:

[tex]\[ L \approx 0.0033 \text{ kg m}^2/\text{s} \][/tex]

Solution:

Let the slope of the best fit line be represented by '[tex]m_{best}[/tex]'

and the slope of the worst fit line be represented by '[tex]m_{worst}[/tex]'

Given that:

[tex]m_{best}[/tex] = 1.35 m/s

[tex]m_{worst}[/tex] = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

[tex]\Delta m = \frac{m_{best}-m_{worst}}{2}[/tex]               (1)

Substituting values in eqn (1), we get

[tex]\Delta m = \frac{1.35 - 1.29}{2}[/tex] = 0.03 m/s

Answer:

9.66 x 10^-6 m

Explanation:

Use the Wein's displacement law

[tex]\lambda _{m}\times T = b[/tex]

Where, b is the Wein's constant

b =  2.898 x 10^-3 meter-kelvin

So, λm x 300 =  2.898 x 10^-3

λm = 9.66 x 10^-6 m

im not that smart but maybe 2 x the mass of the wieght will give u the answer

Answer:

The average force exerted on the bullet are of F=9000 Newtons.

Explanation:

t= 2*10⁻³ s

m= 0.03 kg

V= 600 m/s

F*t= m*V

F= (m*V)/t

F= 9000 N

The average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N). when a bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gunpowder.

Given:

Mass of the bullet (m) = 0.0300 kg

The final velocity of the bullet (v) = 600 m/s

Time taken to reach the final velocity (t) = 2.00 ms = 2.00 × 10⁻³ s

acceleration (a) = (change in velocity) / (time taken)

a = (v - u) / t

a = (600 - 0 ) / (2.00 × 10⁻³)

Now, we can calculate the average force using Newton's second law:

force (F) = mass (m) × acceleration (a)

F = 0.0300 × [(600 ) / (2.00 × 10⁻³)]

F = 0.0300× (3.00 × 10⁵)

F = 9000 N

Therefore, the average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N).

To know more about the force exerted:

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Answer:

decrease

Explanation:

If the two charged ball attracts each other, it means the charge on both the balls are opposite in nature.

As, we insert a glass slab, it means a dielectric is inserted in between the charges. The force between them is reduced.

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance [tex]r= 5.00\ \Omega[/tex]

(a). We need to calculate the current

Using rule of loop

[tex]E-IR-Ir=0[/tex]

[tex]I=\dfrac{E}{R+r}[/tex]

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

[tex]I=\dfrac{3.200}{1.00\times10^{3}+5.00}[/tex]

[tex]I=3.184\times10^{-3}\ A[/tex]

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

[tex]V=E-Ir[/tex]

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

[tex]V=3.200-3.184\times10^{-3}\times5.00[/tex]

[tex]V=3.18\ V[/tex]

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

[tex]\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }[/tex]

[tex]\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}[/tex]

Hence, This is the required solution.

The current flowing in the circuit is 3.195 milliamps. The terminal voltage is calculated to be 2.984V. The ratio of the terminal voltage to the emf is 0.9325 which shows that the terminal voltage is 93.25% of the emf due to the voltage drop resulted from the internal resistance of the battery.

The question is about the terminal voltage and internal resistance of a battery. To answer this, first we need to understand that terminal voltage is the potential difference (voltage) between two terminals of a battery, and it's slightly less than the emf due to internal resistance of the battery.

(a) The current (I) in the circuit is found using Ohm’s law: I = emf / (R_load + r) = 3.200V / (1.00kΩ + 5.00Ω) = 3.195 milliamps.

(b) The terminal voltage (V) is calculated by: V = emf - Ir = 3.200V - (3.195mA * 5.00Ω) = 2.984V.

(c) The ratio of the measured terminal voltage to the emf is V / emf = 2.984V / 3.200V = 0.9325. This shows that the terminal voltage is 93.25% of the emf, which accounts for the voltage drop due to the internal resistance of the battery.

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Answer:

change that a lead is 0.13533

Explanation:

µ  = 50000

f(t) = [e^(-t/µ )]/[µ      if  t ≥ 0

f(t) = 0  if  t < 0

τ = 100000

to find out

the chance that a led will last

solution

we know function is f(t) = [e^(-τ/µ)]/[µ]    

τ = 100000

so we can say that probability (τ  ≥ 100000 ) that is

= 1 - Probability ( τ ≤ 100000 )

that is function of F so

= 1 - f ( 100000 )

that will be

= 1 - ( 1 - [e^(-τ/µ)]/[µ]   )

put all value here τ = 100000 and µ = 50000

= 1 - ( 1 - [e^(-100000/50000)]  )

= 1 - 1 - [e^(-100000/50000)]

= 0.13533

so that change that a lead is 0.13533

Answer:

A typical example of an elastic collision that can be observed is the collision of billiard balls, while an inelastic collision is presented in cars collisions.

Explanation:

In an inelastic collision, the energy system is lost in making the permanent deformation over car's structures due to the impact. As can be stated below, the final and initial kinetic energy are expressed:

[tex] Ei =0.5*m1.v1_{i}^2+m2.v2_{i}^2[/text]  

[tex] Ef =0.5*m1.v1_{f}^2+m2.v2_{f}^2[/text]  

Where the subscripts 1 and 2 relate to each car. In the final energy equation Ef, the car's final velocity will be lower than the respective initial velocities.

[tex] v1_{f}<v1_{i}[/text]  

[tex] v2_{f}<v2_{i}[/text]  

Take into account that car's masses still being the same after the collision, therefore the energy losses are always because of cars velocities changes:

[tex] Ef<Ei[/text]

In the elastic collision, there will be little or negligible deformations and that won't make energy losses. But this statement doesn't affirm that billiard balls velocities will be the same. In fact, could happen that one ball increases its velocities if the other ball decreases its velocity, but taking into account that the energy will always conserve.

[tex] v2_{f}>v2_{i}[/text] if [tex] v1_{f}<v1_{i}[/text]

or  

[tex] v1_{f}>v1_{i}[/text] if [tex] v2_{f}<v2_{i}[/text]

Under the assumption that balls masses still being the same:

[tex] Ef=Ei[/text]

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

[tex]PV=nRT[/tex]

For a gas

[tex]P_{1}V_{1}=nRT_{1}[/tex]

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

[tex]10\times V=nR\times286[/tex]....(I)

When the temperature of the gas is increased

Then,

[tex]P_{2}V_{2}=\dfrac{n}{2}RT_{2}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}[/tex]

[tex]\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}[/tex]

[tex]P_{2}=\dfrac{10\times368}{2\times286}[/tex]

[tex]P_{2}= 6.433\ atm[/tex]

[tex]P_{2}=6.4\ atm[/tex]

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

Final answer:

The radius of the circle is approximately 1.392 m (rounded to one decimal place).

Explanation:

To find the radius of the circle, we need to use the formula for the area of a sector. The area of a sector is given by the formula A = (θ/2π) × πr², where θ is the central angle in radians and r is the radius. In this case, we are given that the central angle is 2π/11 radians and the area is 25 m². We can set up the equation as 25 = (2π/11) × πr² and solve for r.

Solution:

25 = (2π/11) × πr²

25 = (2π²/11) × r²

r² = 11/2π

r ≈ √(11/2π)

r ≈ 1.392 m (rounded to one decimal place)

Answer:

4.49 x 10^-11 newton

Explanation:

v = 12.3 m/s along north = 12.3 j m/s

B = 4.78 x 10^-5 T downwards = 4.78 x 10^-5 k T

q = 7.64 x 10^-8 C

force on a charged particle when it is moving in a uniform magnetic field is given by

F = q (v x B )

F = 7.64 x 10^-8 {(12.3 i) x (4.78 x 10^-5 k)}

F = 4.49 x 10^-11 (- k) newton

magnitude of force = 4.49 x 10^-11 newton

Answer:

[tex]Q = 1.45 \times 10^{-7} C[/tex]

Explanation:

Here flux passing through each surface is given

so total flux through whole cube is given by

[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]

[tex]\phi = 16400 Nm^2 C[/tex]

now we also know that total flux through a closed surface depends on the total charge enclosed in the surface

So we will have

[tex]\frac{Q}{\epsilon_0} = 16400[/tex]

[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]

[tex]Q = 1.45 \times 10^{-7} C[/tex]

The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:

[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]

where:

- [tex]\( \Phi \)[/tex] is the electric flux,

- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,

- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].

For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].

[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]

Now, we can set up an equation using the given values:

[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]

[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]

Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:

[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]

[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]

[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]

So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

Answer:

133280 N

Explanation:

Volume, V = 1 m^3

density of mercury, d = 13.6 x 10^3 kg/m^3

Buoyant force, F = Volume immersed x density of mercury x g

F = 1 x 13.6 x 1000 x 9.8

F = 133280 N

Answer:

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Explanation:

Time period of simple pendulum is given by

         [tex]T=2\pi\sqrt{\frac{l}{g}}[/tex], l is the length of pendulum, g is acceleration due to gravity value.

We can solve acceleration due to gravity as

            [tex]g=\frac{4\pi^2l}{T^2}[/tex]

Here

  Length of pendulum = 1.20 m

  Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.

  Period, [tex]T=\frac{450}{100}=4.5s[/tex]

Substituting

         [tex]g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2[/tex]

Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²

Answer:

0.207 ms

Explanation:

First of all we need to find the length of the pendulum at 20 degrees. We know that the period is 1 s, and the formula for the period is

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]

where L is the length of the pendulum and g is the gravitational acceleration. Solving the equation for L and using T = 1 s and g = 9.8 m/s^2, we find

[tex]L=g(\frac{T}{2\pi})^2=(9.8) (\frac{1}{2\pi})^2=0.248237 m[/tex]

Now we can find the new length of the pendulum at 43 degrees; the coefficient of thermal expansion of brass is

[tex]\alpha =18\cdot 10^{-6} 1/^{\circ}C[/tex]

And the new length of the pendulum is given by

[tex]L' = L (1+\alpha \Delta T)[/tex]

where in this case

[tex]\Delta T = 43-20 = 23^{\circ}[/tex] is the change in temperature

Substituting,

[tex]L'=(0.248237)(1+(18\cdot 10^{-6})(23))=0.248340 m[/tex]

So we can now calculate the new period of the pendulum:

[tex]T'=2\pi \sqrt{\frac{L'}{g}}=2\pi \sqrt{\frac{0.248340}{9.8}}=1.000208 s[/tex]

So the change in the period is

[tex]T'-T=1.000208 - 1.000000 = 0.000207 s = 0.207 ms[/tex]

Final answer:

The period of a pendulum clock with a brass suspension system will change by approximately 0.000414 seconds when the temperature increases from 20°C to 43°C.

Explanation:

A pendulum clock with a brass suspension system is calibrated to have a period of 1 second at 20 degrees Celsius. When the temperature increases to 43 degrees Celsius, the period of the pendulum will change. To calculate the change in period, you can use the formula T2 = T1 * (1 + α * (T2 - T1)), where T2 is the final temperature, T1 is the initial temperature, and α is the coefficient of linear expansion for the brass material. In this case, α is 18 × 10^-6 °C^-1.

Using the formula, we can plug in the values: T1 = 20°C, T2 = 43°C, and α = 18 × 10^-6 °C^-1. Subtracting T1 from T2 gives us 23, and multiplying this by α gives us 0.000414. Finally, multiplying this by the initial period of 1 second gives us a change in period of approximately 0.000414 seconds.