To make a bounce pass, a player throws a 0.60-kg basketball toward the floor. The ball hits the floor with a speed of 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

Answer:

The total current supplied by the source of voltage = 10.29 A

Explanation:

We have a 14-Ω coffee maker and a 14-Ω frying pan are connected in series.

Effective resistance = 14 + 14 = 28Ω

Now we have 28Ω and 20Ω in parallel

Effective resistance

             [tex]R=\frac{28\times 20}{28+20}=11.67\Omega[/tex]

So we have resistor with 11.67Ω in a 120 V source of voltage.

We have equation V = IR

Substituting

               120 = I x 11.67

                 I = 10.29 A

The total current supplied by the source of voltage = 10.29 A

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:

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:

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:

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]

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:

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:

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:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

        Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

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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The orbital velocity of the satellite can be calculated using the formula v = √(GM/r), where v is the orbital velocity, G is the universal gravitational constant, M is the mass of Mars, and r is the radius of the orbit.

To calculate the orbital velocity of the satellite, we can use the formula for orbital velocity:

v = √(GM/r)

where v is the orbital velocity, G is the universal gravitational constant, M is the mass of Mars, and r is the radius of the orbit. Plugging in the known values, we have:

v = √((6.674×10-11 m3/kg/s2)(6.419×1023 kg)/(3396000 m + 300000 m))

Calculating this will give us the orbital velocity of the satellite.

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The orbital velocity of the satellite is approximately 3,404 meters per second.

Sure, here is the solution to the problem:

Given:

Radius of Mars (R) = 3,396 km = 3.396 × 10⁶ m

Mass of Mars (M) = 6.419 × 10²³ kg

Universal gravitational constant (G) = 6.674 × 10⁻¹¹ m³/kg·s²

To find:

Orbital velocity (v)

Formula:

The orbital velocity of a satellite in a circular orbit is given by the following formula:

v = √(GM / r)

where:

G is the universal gravitational constant

M is the mass of the planet

r is the radius of the orbit

Calculation:

First, convert the radius of Mars from kilometers to meters:

r = 3.396 × 10⁶ m

Now, plug in the values into the formula:

v = √((6.674 × 10⁻¹¹ m³/kg·s²) × (6.419 × 10²³ kg) / (3.396 × 10⁶ m))

v ≈ 3,404 m/s

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]

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:

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.

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:

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

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

Given:

applied tensile stress, [tex]\sigma[/tex] = 170 MPa

radius of curvature of crack tip,  [tex]r_{t}[/tex] =  [tex]3.5\times 10^{-4}[/tex] mm

crack length = [tex]4.5\times 10^{-2}[/tex] mm

half of internal crack length, a = [tex]\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}[/tex]

a =  [tex]2.25\times 10^{-2}[/tex]

Formula Used:

[tex]\sigma _{max} =  2\times\sigma \sqrt{\frac{a}{r_t}}[/tex]

Solution:

Using the given formula:

[tex]\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}[/tex]

[tex]\sigma _{max}[/tex] = 2726 MPa (395372.9 psi)

The magnitude of the maximum stress at the tip of an internal crack can be determined using the stress concentration factor formula.

The magnitude of the maximum stress at the tip of an internal crack can be determined using the formula for the stress concentration factor, which is the ratio of the maximum stress to the applied stress. The stress concentration factor (Kt) for an internal crack can be calculated using the following equation:

Where Kt is the stress concentration factor, a is the crack length, and r is the radius of curvature of the crack.

Plugging in the given values:

Kt = 1 + 2 * (4.5 x 10-2 mm) / (3.5 x 10-4 mm) = 127

The magnitude of the maximum stress can be calculated by multiplying the stress concentration factor by the applied stress:

Maximum Stress = 127 * 170 MPa = 21,590 MPa

The reactance of an inductor is given by:

X = 2πfL

X is the inductor's reactance

f is the frequency of the supplied voltage

L is the inductor's inductance

The given values are:

f = 60.0Hz

L = 43.8mH (I'm assuming the value is given in milli Henries because this is within the normal range of inductors)

Plug these values in and solve for X:

X = 2π(60.0)(43.8×10⁻³)

X = 16.512Ω

Round this value to 3 significant figures:

X = 16.5Ω

The relationship between AC voltage and current is given by:

V = IZ

V is the voltage

I is the current

Z is the impedance

For an AC inductor circuit, Z = X = 16.512Ω and V is the rms voltage 120V. Plug these values in to get the rms current:

120 = I×16.512

I = 7.2673A

Round this value to 3 significant figures:

I = 7.27A

Final answer:

The inductive reactance is 16,515 Ohms and the rms current through the inductor is 7.3 mA for an AC source with an rms voltage of 120 V operating at a frequency of 60 Hz.

Explanation:

To determine the inductive reactance and the rms current through the inductor in a purely inductive AC circuit, we use the inductive reactance formula XL = 2πfL, where f is the frequency and L is the inductance of the coil.

In this case, the frequency f is 60 Hz and the inductor has an inductance L of 43.8 H. The inductive reactance, XL, can be calculated as:

XL = 2π × 60 Hz × 43.8 H ≈ 16,515 Ohms (or 16.5 kΩ)

Once we have the inductive reactance, we can calculate the rms current using Ohm's law, I = V/XL, where I is the current and V is the rms voltage of the AC source. With an rms voltage of 120 V, the rms current is:

I = 120 V / 16,515 Ohms ≈ 0.0073 A (or 7.3 mA)

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.

Answer:

a) The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation, s = 17.3dB

Explanation:

Noise levels at 5 airports = 147,123,119,161,136

Mean noise level

        [tex]\bar{x} =\frac{ 147+123+119+161+136}{5}=137.2dB[/tex]

Variance of noise level

        [tex]\sigma^2 =\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5}\\\\\sigma^2=164.16[/tex]

Standard deviation,

       [tex]\sigma =\sqrt{164.16}=12.81dB[/tex]

a) Confidence interval  is given by

       [tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]

For 99% confidence interval Z = 2.576,

Number of noises, n = 5

Substituting

    [tex]137.2-2.576\times \frac{12.81}{\sqrt{5}}\leq \mu\leq 137.2+2.576\times \frac{12.81}{\sqrt{5}}\\\\122.44\leq \mu\leq 151.96[/tex]

The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation

        [tex]s=\sqrt{\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5-1}}\\\\s=17.3dB[/tex]

   Sample standard deviation, s = 17.3dB

Answer:

5.6 Tesla

Explanation:

L = 52 cm = 0.52 m

V = 2.75 m/s

e = 8 V

Let B be tha magnitude of magnetic field. Use the formula for the motional emf

e = B × V × L

B = e / V L

B = 8 / (2.75 × 0.52)

B = 5.6 Tesla

Answer:

1.13 Wb

Explanation:

First of all, we need to find the area enclosed by the coil.

The perimeter of the square is 2.0 m, so the length of each side is

[tex]L=\frac{2.0}{4}=0.5 m[/tex]

So the area enclosed by the coil is

[tex]A=L^2 = (0.5 m)^2=0.25 m^2[/tex]

Now we can calculate the magnetic flux through the square, which is given by

[tex]\Phi = B A cos \theta[/tex]

where

B = 9.0 T is the strength of the magnetic field

[tex]A=0.25 m^2[/tex] is the area of the coil

[tex]\theta[/tex] is the angle between the direction of the magnetic field and the normal to the coil; since the field is oriented 30.0° above the horizontal and the coil lies in the horizontal plane, the angle between the direction of the magnetic field and the normal to the coil is

[tex]\theta=90^{\circ}-30^{\circ}=60^{\circ}[/tex]

So the magnetic flux is

[tex]\Phi = (9.0)(0.25)(cos 60^{\circ})=1.13 Wb[/tex]

The magnetic flux through the square is 18.0 T·m²

To find the magnetic flux through the square, we need to calculate the area of the square and the component of the magnetic field perpendicular to the square's plane.

The area of the square is given by A = (side length) = (2.0 m)² = 4.0 m²

The component of the magnetic field perpendicular to the square's plane is B_perpendicular = B × sin(30°) = 9.0 T × sin(30°) = 4.5 T.

Therefore, the magnetic flux through the square is given by the product of the area and the component of the magnetic field perpendicular to the square's plane: flux = B_perpendicular × A = 4.5 T × 4.0 m²= 18.0 T·m²

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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:

The pilot must keep the tip pointed at 12.1 degrees to the right with respect to the direction of the runway to align the flight path with the runway.

Explanation:

x= -10m/s

y= 47m/s

r= √(x²)+(y²)

r=48.05 m/s

β= tan⁻¹(y/x)

β=102.01°

the runway is at 90 degrees. Considering the wind, the airplane is flying at 102.01 º direction. Must fly at 12.1 degrees to the right with respect to the direction of the runway to contrarest the wind effect.

The launch angle that results in the maximum range of a projectile up an incline depends on the initial speed and the angle of the incline. For conditions neglecting air resistance, the maximum range is obtained at 45 degrees. If air resistance is considered, the maximum angle is around 38 degrees.

The range of a projectile launched up an incline depends on the launch angle. To determine the launch angle that results in the maximum range, we need to consider the initial speed and the angle of the incline. Figure 3.38(b) shows that for a fixed initial speed, the maximum range is obtained at 45 degrees. However, this is only true for conditions neglecting air resistance. If air resistance is considered, the maximum angle is around 38 degrees. It is also interesting to note that for every initial angle except 45 degrees, there are two angles that give the same range, and the sum of those angles is 90 degrees.

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.

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]