A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?

Answer:

I don't understand?

Explanation:

Could you please elaborate?

Answer:

Mass of the sphere is 19.2 kg

Explanation:

We have given diameter of the sphere d = 0.74 m

So radius r = 0.37 m

Initial angular velocity [tex]\omega _i=0rad/sec[/tex]

Time t = 14 sec

Angular displacement [tex]\Theta =160revolution=160\times 2\pi =1004.8rad[/tex]

From second equation of motion

[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]

So [tex]1004.8=0\times t+\frac{1}{2}\times \alpha \times 14^2[/tex]

[tex]\alpha =10.25rad/sec^2[/tex]

Torque is given [tex]\tau =10.8Nm[/tex]

Torque is equal to [tex]\tau =I\alpha[/tex], here I is moment of inertia and [tex]\alpha[/tex] angular acceleration

So [tex]10.8=10.25\times I[/tex]

[tex]I=1.053kgm^2[/tex]

Moment of inertia of sphere is equal to [tex]I=\frac{2}{5}Mr^2[/tex]

So [tex]1.053=\frac{2}{5}\times M\times 0.37^2[/tex]

M = 19.23 kg

So mass of the sphere is 19.23 kg

Answer:2940

Explanation:

Answer:

0.635 kg

Explanation:

Recall that Kinetic energy is defined as:

K.E. = (1/2) mv²,

where

K.E = Kinetic energy = given as 140J

m = mass (we are asked to find this)

v = velocity = given as 21 m/s

Substituting the above values into the equation:

K.E. = (1/2) mv²

140 = (1/2) m (21)²   (rearranging)

m = (140)(2) / (21)²

m = 0.635 kg

Answer:

0.4455 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

Total mass is

[tex]m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg[/tex]

Here the spring constant is not given so let us assume it as [tex]k=36000\ N/m[/tex]

Here, the forces are balanced

[tex]mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m[/tex]

The springs are compressed by 0.4455 m

Answer:

Gravity

Explanation:

I know because im a verified expert

Answer:

gravity

Explanation:

1. gravity is the force that pulls us to the surface of the Earth, keeps the planets in orbit around the Sun and causes the formation of planets, stars and galaxies.

2. electromagnetism is the force responsible for the way matter generates and responds to electricity and magnetism.

Answer:

flash

Explanation:

Answer:

Sonic obviously

Explanation:

Sonic's body was made to run, the flashes is just a normal human body.

Answer:

86.48 dB . Ans

Explanation:

Sound level by one singer = 70.8 dB

= 7.08 B

If I be its intensity in terms of watt/m²

log ( I / 10⁻¹² ) = 7.08 ( given )

I / 10⁻¹²  = 10⁷°⁰⁸

I = 10⁻¹² x 10⁷°⁰⁸

= 10⁻⁴°⁹²

= .000012022 W / m²

intensity by 37 singers

= 37 x .000012022 W /m²

= .0004448378

= 4448.378 x 10⁻⁷

intensity level

= log I / 10⁻¹² B

= log 4448.378 x 10⁻⁷ / 10⁻¹²

= log 444837800 B

log 4448378 + 2 B

= 6.6482 +2 B

8.6482 B

intensity level = 8.6482 B

= 86.48 dB . Ans

Answer:

86.48 dB

Explanation:

86.48 dB

Explanation:

sound level intensity, β = 70.8 dB  

Io = 10^-12 W/m²

Let the intensity if I at the level of 70.8 dB  

Use the formula  

7.08 = log I + 12

log I = - 4.92  

I = 1.2 x 10^-5 W/m²

There are 37 singers, so the total intensity is I'.  

I' = 37 x I  

I' = 37 x 1.2 x 10^-5 = 4.45 x 10^-4 W/m²

Let the sound intensity is β' in decibels.  

Use the formula  

β' = 86.48 dB

Answer:

Spectrophotometry is the quantitative measurement of light spectra reflection and transmission properties of materials as function of the wavelength. Photometry measures the total brightness as seen by the human eye

Explanation:

Please give me brainliest

Answer:

a) 2.90*10^-6 T

b) 0.092A

Explanation:

a) The magnitude of the magnetic field is given by the formula for the calculation of B when it makes an electron moves in a circular motion:

[tex]B=\frac{m_ev}{qR}[/tex]

me: mass of the electron = 9.1*10^{-31}kg

q: charge of the electron = 1.6*10^{-19}C

R: radius of the circular path = 4.3cm=0.043m

v: speed of the electron = 2.20*10^4 m/s

By replacing all these values you obtain:

[tex]B=\frac{(9.1*10^{-31}kg)(2.20*10^4 m/s)}{(1.6*10^{-19}C)(0.043m)}=2.90*10^{-6}T=2.9\mu T[/tex]

b) The current in the solenoid is given by:

[tex]I=\frac{B}{\mu_0 N}=\frac{2.90*10^{-6}T}{(4\pi*10^{-7}T/A)(25)}=0.092A=92mA[/tex]

To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid. To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field.

To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid: B = µ0nI, where B is the magnetic field, µ0 is the permeability of free space (4π × 10-7 T·m/A), n is the number of turns per unit length, and I is the current flowing through the solenoid.

Given that the solenoid has 25 turns per centimeter and the radius of the circular path is 4.3 cm, we can find the number of turns per unit length: n = 25 × 100 cm = 2500 turns/m.

The strength of the magnetic field inside the solenoid is then: B = (4π × 10-7 T·m/A) × (2500 turns/m) × (I).

To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field: F = qvB, where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field.

Given that the velocity of the electron is 2.20 × 104 m/s and the radius of the circular path is 4.3 cm, we can find the magnetic force experienced by the electron: F = (1.6 × 10-19 C) × (2.20 × 104 m/s) × (B).

Since the magnetic force is centripetal, it can also be expressed as F = mv2/r, where m is the mass of the electron, v is its velocity, and r is the radius of the circular path.

Combining these two equations, we can find the current in the solenoid: I = (mv2/r) / (vB).

The third point charge must be located at the point where the net electric force on it is zero.

The net electric force on a point charge is the vector sum of the electric forces exerted on it by all the other point charges. The electric force exerted by a point charge q1 on a point charge q2 is given by Coulomb's law:

F = k * |q1| * |q2| / r^2

where k is Coulomb's constant, |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.

The direction of force will be opposite for both charges, and since net force is zero, the force due to both charges will be equal and opposite.

So, |q1| * (1/r1^2) = |q2| * (1/r2^2)

Where r1 is the distance between point charge 1 and point charge 3, r2 is the distance between point charge 2 and point charge 3.

Solving for position of point charge 3, we can say that it is located at y = 0.

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To find the position where a third charge experiences zero net force due to two other charges along the y-axis, we must consider the magnitudes and signs of all charges and use Coulomb's Law. The third charge will be or negative and situated closer to the smaller magnitude charge to balance the forces. The exact position, however, cannot be determined without the magnitude of the third charge.

The question pertains to the concept of electric forces and charge equilibrium in Physics. We are asked to find the position where a third charge would experience a net electric force of zero due to the presence of two other point charges arranged along the y-axis. The charges are q1 = − 9.0 μC at y = 6.0 m and q2 = − 8.0 μC at y = − 4.0 m.

Let's denote the position of the third charge as y3 where the net force is zero. To ensure a net force of zero on the third charge, the electric force due to q1 must be equal in magnitude and opposite in direction to the force due to q2. Since both q1 and q2 are negative, the third charge must also be negative (so it will be repelled by both). The third charge must be placed between q1 and q2.

Likewise, since q1 is larger in magnitude than q2, the third charge must be placed closer to q2 to balance the forces (since electric force is inversely related to the square of the distance). Mathematically, we can set up the equation where the magnitude of the forces due to each charge on the third charge is equal. This investigation involves Coulomb's Law (F = k · |q1 · q3| / r^2), where F is the force between charges, k is Coulomb's constant, q3 is the third charge, and r is the distance between the charges involved.

Through setting up the equations and solving for y3, we can determine the precise location. Unfortunately, the problem does not provide the magnitude of the third charge, so while we can describe the process, we can't compute a specific answer without this information. In a real scenario though, any negative third charge would suffice, and its magnitude would not affect the balance point.

The magnitude of the angular velocity immediately after the collision is

[tex]\( 14.4 \, \text{rad/s} \)[/tex].

To find the angular velocity immediately after the collision, we can apply the principle of conservation of angular momentum. The angular momentum before the collision equals the angular momentum after the collision.

The angular momentum before the collision is zero since the rod is at rest. After the collision, the bullet-rod system rotates together. We can calculate the moment of inertia [tex]\( I \)[/tex] of the system and then use the equation [tex]\( L = I\omega \)[/tex], where [tex]\( L \)[/tex] is the angular momentum and [tex]\( \omega \)[/tex] is the angular velocity.

First, calculate the moment of inertia of the system:

[tex]\[ I = \frac{1}{3}mL^2 + md^2 \][/tex]

[tex]\[ I = \frac{1}{3}(3)(0.25)^2 + (0.0045)(0.25/4)^2 \][/tex]

[tex]\[ I = 0.015625 \, \text{kg m}^2 \][/tex]

Now, apply conservation of angular momentum:

[tex]\[ I\omega = mvd \][/tex]

[tex]\[ \omega = \frac{mvd}{I} \][/tex]

[tex]\[ \omega = \frac{(0.0045)(300)(0.25/4)}{0.015625} \][/tex]

[tex]\[ \omega = 14.4 \, \text{rad/s} \][/tex]

So, the magnitude of the angular velocity immediately after the collision is [tex]\( 14.4 \, \text{rad/s} \)[/tex].

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis? (d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?

Given Information:

Radius of parallel-plate capacitor = 4.10 cm = 0.041 m

Conduction current = Ic = 0.276 A

Required Information:

a) Displacement current density = JD = ?

b) Rate of change of electric field = dE/dt = ?

c) Magnetic field between plates at r = 2 cm = ?

d) Magnetic field between plates at r = 1 cm = ?

Answer:

a) Displacement current density = JD = 52.27 A/m²

b) Rate of change of electric field = dE/dt = 5.904×10¹² V/m.s

c) Magnetic field between plates at r = 2 cm = 6.568×10⁻⁷ Tesla

d) Magnetic field between plates at r = 1 cm = 3.284×10⁻⁷ Tesla

Explanation:

a) What is the displacement current density JD in the air space between the plates?

Displacement current density is given by

JD = Id/A

Where Id is the conduction current and A is the area of capacitor given by

A = πr²

A = π(0.041)²

A = 0.00528 m²

As you can notice in the diagram, conduction current has equal displacement between the capacitor plates therefore, Id = Ic

JD = 0.276/0.00528

JD = 52.27 A/m²

b) What is the rate at which the electric field between the plates is changing?

The rate of change of electric field is given by

dE/dt = JD/ε₀

Where JD is the displacement current density and ε₀ is the permittivity of free space and its value is 8.854×10⁻¹² C²/N.m²

dE/dt = 52.27/8.854×10⁻¹²

dE/dt = 5.904×10¹² V/m.s

c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis?

The induced magnetic field between the plates can be found using Ampere's law

B = (μ₀/2)*JD*r

Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷ T.m/A, JD is the displacement current density and r is the distance from the axis.

B = (4π×10⁻⁷/2)*52.27*0.02

B = 6.568×10⁻⁷ Tesla

d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?

B = (μ₀/2)*JD*r

B = (4π×10⁻⁷/2)*52.27*0.01

B = 3.284×10⁻⁷ Tesla

This physics-based question involves understanding the properties and behaviors of a parallel-plate, air-filled capacitor. Specifically, it focuses on finding the displacement current density and the rate at which the electric field between the plates is changing as the capacitor is charged.

The question revolves around a parallel-plate capacitor, a system of two identical conducting plates separated by a distance. This specific case involves an air-filled capacitor where the plates have a certain radius, and a conduction current is defined.

First, let's address part (a): What is the displacement current density jD in the air space between the plates? The displacement current density, noted as 'jD', can be found by using Maxwell's equation which correlates the time rate of change of electric flux through a surface to the displacement current passing through the same surface.

For (b): What is the rate at which the electric field between the plates is changing? As the current is introducing charge onto the plates, the electric field (which is proportional to the charge on the capacitor's plates) will be changing. This change can be calculated as the derivative of the electric field's magnitude with respect to time.

Without specific numerical values or more information, I'm unable to provide a more detailed analysis or a concrete answer.

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Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

     Q_lost = - Q_absorbed

The latent heat is

      Q_absorbed = m L

The heat given by the body

      Q_lost = M [tex]c_{e}[/tex] ΔT

where m is the mass of sweat and M is the mass of the body

       m L = M c_{e} ΔT

        m = M c_{e} ΔT / L

let's replace

        m = 90  3.500  1.8 / 2.42 10⁶

        m = 0.2343 kg

reduced to grams

        m = 0.2342 kg (1000g / 1kg)

        m = 23.42 g

 the correct answer is c) 23 g

230 g should have to evaporate.

Given that,

The calculation is as follows:

[tex]= \frac{(90kg)(3500J/kg^{\circ})(1.8^{\circ}C)}{2.42\times 10^6J/kg}[/tex]

= 0.23 kg

= 230g

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Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

[tex]r=2.3*sin70=2.161m[/tex] (ball rotates in a circle)

If the system is in equilibrium, the tension is:

[tex]Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}[/tex]

Replacing:

[tex]\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}[/tex]

Replacing:

[tex]v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s[/tex]

Answer:

[tex]0.293I_0[/tex]

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

[tex]I_1=\frac{I_0}{2}[/tex]

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

[tex]I_2=I_1 cos^2 \theta[/tex]

where

[tex]\theta[/tex] is the angle between the axes of the two polarizer

Here we have

[tex]\theta=40^{\circ}[/tex]

So the intensity after the 2nd polarizer is

[tex]I_2=I_1 (cos 40^{\circ})^2=0.587I_1[/tex]

And substituting the expression for I1, we find:

[tex]I_2=0.587 (\frac{I_0}{2})=0.293I_0[/tex]

Answer:

Explanation:

Given that,

His friend design has a turnable disk of radius 1.5m

R = 1.5m

The mass is twice the fully loaded elevator.

Let the mass of the full loaded elevator be M

Then, mass of the turn able

Mt = ½M

Radius of the disk that serves as a vertical pulley is ¼ radius of turntable and 1/16 of the mass.

Mass of pulley is

Mp = 1 / 16 × Mt

Mp = 1 / 16 × M / 2

Mp = M / 32

Also, radius of pulley

Rp = ¼ × R = ¼ × 1.5

Rp = 0.375m

The moment of inertia of the disk of a ring is

I = ½MR²

To calculate the moment of the turntable, we can use the formula

I_t = ½Mt•R²

I_t = ½ × ½M × 1.5²

I_t = 0.5625 M

I_t = 9M / 16

Also, the moment of inertia of the vertical pulley

I_p = ½Mp•Rp

I_p = ½ × (M/16) × 0.375

I_p = 0.01171875 M

I_p = 3M / 256

Let assume that, the tension in the cable between the pulley and the elevator is T1 and Let T2 be the tension between the turntable and the pulley

So, applying newton second law of motion,

For the elevator

Fnet = ma

Mg - T1 = Ma

a = (Mg-T1) / M

For vertical pulley,

The torque is given as

τ_p = I_p × α_p = (T2—T1)•r

τ_p = 3M/256 × α_p = (T2-T1)•r

For turntable

The torque is given as

τ_t = I_t × α_t = T2•r

τ_t = 9M/16 × α_t = T2•r

So, the torque are equal

τ_t = τ_p

9M/16 × α_t = 3M/256 × α_p

M cancel out

9•α_t / 16 = 3•α_p / 256

Cross multiply

9•α_t × 256 = 3•α_p × 16

Divide both sides by 48

48•α_t = α_p

α_t = α_p / 48

Then, from,

τ_t = 9M/16 × α_t = T2•r

T2•r = 9M / 16 × α_p / 48

T2•r = 3Mα_p / 16

Also, from

τ_p = 3M/256 × α_p = (T2-T1)•r

3M/256 × α_p = T2•r - T1•r

T1•r = T2•r - 3M/256 × α_p

T1•r = 3Mα_p / 16 - 3M/256 × α_p

T1•r = 3Mα_p / 16 - 3Mα_p/256

T1•r = 45Mα_p / 256

T1 = 45Mα_p / 256R

Then, from

a = (Mg-T1) / M

a = Mg - (45Mα_p / 256R) / M

a = g - 45α_p / 256

From the final answer, it is show that the acceleration is always less than acceleration due to gravity due to the subtraction of 45α_p / 256 from g

Answer:

Explanation:

Given that,

The light starts at a point 8cm beneath the surface

It strikes the surface 7cm directly above the light

The angle of incidence from the given distances is

Using trigonometry

Tan I = opp / adj

tan I= (7.0 cm)/(8.0 cm) = 0.875

I = arctan(0.875)

I = 41.1859

According to Snell's law

n(water) sin I = n(air) sin R

Here R =90° for total reflection to occur

nair = 1

n(water) sin I = n(air) sinR

n(water) = (1)sin 90

n(water) = 1

n(water) = 1/sin I

n(water) =1/ sin(41.1859o)

= 1.52

The refractive index index of the liquid is 1.52

To find the index of refraction of the liquid, we calculate the angle of incidence using the given distances and apply Snell's law. While specific numeric calculations aren't provided, the assumption is the liquid's refractive index is greater than air, as total internal reflection occurs.

A beam of light undergoes total internal reflection when it moves from a medium of higher refractive index to one of lower refractive index, and the angle of incidence is greater than the critical angle. In this case, we are given that a beam of light in a liquid undergoes total internal reflection. To determine the index of refraction of the liquid, we can use the information given about the distances involved to find the angle of incidence and then apply Snell's law.

First, we use the distances given to find the angle of incidence. The light source is 8.0 cm beneath the surface and strikes the surface 7.0 cm from the point directly above it. This forms a right triangle, allowing us to calculate the angle of incidence using trigonometry. The tangent of the angle of incidence (θ) is opposite over adjacent, or θ = tan⁻¹(8.0/7.0). The angle of incidence calculated is therefore tan⁻¹(8/7).

For total internal reflection to occur, the angle of incidence must be greater than the critical angle, which is defined by Snell's law as sin⁻¹(n₂/n₁) where n₂<n₁. Since we know the reflection happens at the interface with air (where n₂ = 1.00 for air), we can find the refractive index of the liquid. However, without exact values for angles given in this scenario, the specific calculation steps cannot be completed, but we understand that the liquid's refractive index is greater than 1, which is typical for most transparent materials compared to air.

Answer:

Secondary voltage of transformer is 905.23 volt  

Explanation:

It is given number of turns in primary of transformer [tex]N_1=275[/tex]

Number of turns in secondary [tex]N_2=2200[/tex]

Input voltage equation of the transformer

[tex]\Delta v=160sin\omega t[/tex]

Here [tex]v_{max}=160volt[/tex]

[tex]v_{rms}=\frac{160}{\sqrt{2}}=113.15volt[/tex]

For transformer we know that

[tex]\frac{V_1}{V_2}=\frac{N_1}{N_2}[/tex]

[tex]\frac{113.15}{V_2}=\frac{275}{2200}[/tex]

[tex]V_2=905.23Volt[/tex]

Therefore secondary voltage of transformer is 905.23 volt

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

[tex]B=\mu_0\, \frac{N}{L} I[/tex]

Then, if we assign the subindex "1" to the quantities that define the magnetic field ([tex]B_1[/tex]) inside solenoid 1, we have:

[tex]B_1=\mu_0\, \frac{N_1}{L_1} I_1[/tex]

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then [tex]L_2=L_1/2[/tex]

2) twice as many turns as solenoid 1. Then [tex]N_2=2\,N_1[/tex]

3) three times the current of solenoid 1. Then [tex]I_2=3\,I_1[/tex]

we obtain:

[tex]B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1[/tex]

The magnetic field at the center of solenoid 2 (B2) will be twelve times larger than the magnetic field at the center of solenoid 1 (B1), due to having four times the number of turns per unit length and three times the current.

The magnetic field inside a solenoid is given by the formula B = µnI, where B is the magnetic field, µ (mu) is the magnetic permeability of the medium, n is the number of turns per unit length, and I is the current through the solenoid. Given solenoid 2 has twice the diameter of solenoid 1, half the length, and twice as many turns, with the current being three times that of solenoid 1, several factors will influence the magnetic field in solenoid 2 (B2) compared to solenoid 1 (B1).

The number of turns per unit length for solenoid 2 is four times that of solenoid 1, since it has twice as many turns and half the length. Additionally, the current in solenoid 2 is three times that in solenoid 1. Therefore, B2 will be twelve times larger than B1, since the magnetic field inside a solenoid is directly proportional to both the number of turns per unit length of the solenoid and the current through it (B2 = 12 * B1).

Answer:

v = 2 m/s

Explanation:

The object moves 20 m in 10 s initially. Then it remains at rest for 20 s. It is required to find the average speed of the object.

Fro 10 s it covers 20 m. For that time, speed is :

[tex]v=\dfrac{d}{t}\\\\v=\dfrac{20\ m}{10\ s}\\\\v=2\ m/s[/tex]

For 20 s, it was at rest means distance covered is 0. So, speed is 0.

The average speed of the object is equal to the sum of speed for 10 s and for 20 s i.e.

v = 2 m/s + 0 = 2 m/s

So, the average speed of the object is 2 m/s.

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

Answer:

[tex]\theta = 52 rads[/tex]

Explanation:

The rotational kinetic equation is given by the formula:

[tex]\theta = w_{0} t + 0.5 \alpha t^{2}[/tex]..............(1)

The object is starting from rest, angular speed, w = 0 rad/s

At t = 2.5 sec, angular displacement, θ = 13 rad

Inserting these parameters into equation (1)

[tex]13 = (0 * 2.5) + 0.5 \alpha 2.5^{2}\\\alpha = \frac{13}{3.125} \\\alpha = 4.16 rad/s^{2}[/tex]

At time, t = 5.0 sec, we substitute the value of [tex]\alpha[/tex] into the kinematic equation in (1)

[tex]\theta = (0*5) + 0.5 *4.16* 5^{2}\\\theta = 52 rad[/tex]

Answer:

[tex]\Delta V = 1.8 \times 10^7 V[/tex]

Explanation:

GIVEN

diameter = 15 fm  =[tex]15 \times 10^{-15}[/tex]m

we use here energy conservation

[tex]K_{i}+U_{i} =K_{f}+U_{f}[/tex]

there will be some initial kinetic  energy but after collision kinetic energy will zero

[tex]K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}[/tex]

on solving these equations we get kinetic energy initial

[tex]KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}[/tex]

[tex]KE_{i} = 35.33[/tex] J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the [tex]{238}_U[/tex] nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

[tex]\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V[/tex]

Answer:

a) [tex]\Delta m = 330000\,kg[/tex], b) [tex]h = 1.268\,m[/tex]

Explanation:

a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:

[tex]\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V[/tex]

[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]

[tex]\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)[/tex]

[tex]\Delta m = 330000\,kg[/tex]

b) The submersion height is found by using the equation derived previously:

[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]

[tex]450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)[/tex]

The final submersion height is:

[tex]h = 1.268\,m[/tex]

Answer:

a) m = 330000 kg = 330 tons

b) H3 = 1.268 meters

Explanation:

Given:-

- The density of fresh-water, ρ = 1000 kg/m^3

- The base area of the rectangular prism boat, A = 550 m^2

- The height of the boat, H = 2.0 m ( empty )

- The bottom of boat barge is H1 = 0.45 m of the total height H under water. ( empty )

- The bottom of boat barge is H2 = 1.05 m of the total height H under water

Find:-

a) Find the mass of the coal in kilograms.

b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge?

Solution:-

- We will consider the boat as our system with mass ( M ). The weight of the boat "Wb" acts downward while there is an upward force exerted by the body of water ( Volume ) displaced by the boat called buoyant force (Fb):

- We will apply the Newton's equilibrium condition on the boat:

                              Fnet = 0

                              Fb - Wb = 0

                              Fb = Wb

Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V1 ). The expression of buoyant force (Fb) is given as:

                             Fb = ρ*V1*g

Where,

                V1 : Volume displaced when the boat is empty and the barge of the boat is H1 = 0.45 m under the water:

                             V1 = A*H1

Hence,

                             Fb = ρ*A*g*H1

Therefore, the equilibrium equation becomes:

                            ρ*A*g*H1 = M*g

                            M =  ρ*A*H1

- Similarly, apply the Newton's equilibrium condition on the boat + coal:

                            Fnet = 0

                            Fb - Wb - Wc = 0

                            Fb = Wb + Wc

Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V2 ). The expression of buoyant force (Fb) is given as:

                             Fb = ρ*V2*g

Where,

                V2 : Volume displaced when the boat is filled with coal and the barge of the boat is H2 = 1.05 m under the water:

                             V2 = A*H2

Hence,

                             Fb = ρ*A*g*H2

Therefore, the equilibrium equation becomes:

                            ρ*A*g*H2 = g*( M + m )

                            m+M = ρ*A*H2

                            m = ρ*A*H2 - ρ*A*H1

                            m = ρ*A*( H2 - H1 )

                            m = 1000*550*(1.05-0.45)

                            m = 330000 kg = 330 tons

- We will set the new depth of barge under water as H3, if we were to add a mass of coal m = 450,000 kg then what would be the new depth of coal H3.

- We will use the previously derived result:

                          m = ρ*A*( H3 - H1 )  

                          H3 = m/ρ*A + H1

                          H3 = (450000 / 1000*550) + 0.45

                          H3 = 1.268 m

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

         I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

          I₁ = I₀ / 2

          I₁ = 845/2

          I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

      Cos² θ = I / I₁

      Cos θ = √ I / I₁

      Cos θ = √ (275 / 422.5)

     Cos θ = 0.80678

     θ = cos⁻¹ 0.80678

     θ = 36.2º

This is the angle between the two polarizers

Answer:

The magnetic field exerts net force and net torque on the loop.

Answer:

n = 1.27

Explanation:

just took test :)

Answer:

N= 1.27

Explanation:

The question is not clear enough. So i have attached a copy of the correct question.

Answer:

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

The magnitude of the magnetic field B is calculated using Faraday's Law and is equal to V/-c. For a square loop being pulled out from the field, the rate of change of area is proportional to the speed of withdrawal and the length of the side still in the field. Hence, c = Lv.

Part A: The magnitude of the magnetic field B can be calculated using Faraday's Law of electromagnetic induction, which states that the induced voltage in a circuit is equal to the negative rate of change of magnetic flux. Mathematically, it is expressed as V = -dB/dt, where V is the induced emf, and dB/dt is the rate of change of magnetic field. Because the area is decreasing at a constant rate, the magnetic field B is V/-c.

Part B: The rate of change of the area, denoted as c = -da/dt, for a square loop of side length L being pulled out of the magnetic field with constant speed v, could be calculated using the formula c = Lv. This is because as the square loop is pulled out with a constant velocity, the rate at which its area decreases is proportional to its speed and the length of the side that is still in the field.

https://brainly.com/question/36936308

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Using conservation of momentum, the lighter piece in the explosion scenario lands 53 meters away from the launch point in the opposite direction in which the heavier piece landed.

The question revolves around a scenario where conservation of momentum applies. An object of mass
3M explodes into two pieces of mass M and 2M. The conservation of momentum dictates that the center of mass of the two pieces and the original object all follow the same horizontal trajectory. If the heavier piece lands 53 meters away from the point of launch, then, due to the center of mass being conserved during the explosion, the lighter piece would have landed at the same point of launch.

Utilizing the conservation law, which states that the total momentum before the explosion must be equal to the total momentum after the explosion, the pieces must move symmetrically about the center of mass of the original object. Given that the larger piece is observed to land 53 meters away from the launch point and has twice the mass of the smaller piece, the smaller piece would have landed 53 meters on the other side of the launch point, making a total separation of 106 meters between the two pieces. Therefore, the lighter piece would have landed 53 meters away in the opposite direction from where the heavier piece landed.

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

Final answer:

To find the displacement that the physics professor must give the car, we can break down the different stages of the car's motion and use the equations of motion. By adding the displacements from each stage, we can find the total displacement of the car.

Explanation:

To find the displacement that the physics professor must give the car in order for its average velocity to be +6.50 m/s for the entire trip, we can break down the different stages of the car's motion and use the equations of motion.

1. From time t=0 to t=3.60s: The car is accelerated in the +x direction at 5.35 m/s². We can use the equation v = u + at to find the final velocity v1 at t=3.60s, where u is the initial velocity, a is the acceleration, and t is the time.

2. From time t=3.60s to t=8.10s: The car slows down at a rate of 1.95 m/s² due to friction. We can use the equation v = u + at to find the initial velocity u2 at t=3.60s, where v is the final velocity, a is the acceleration, and t is the time.

3. From time t=8.10s to t=12.60s: The physics professor pushes the car with a constant velocity for 4.50s. The average velocity during this time is +6.50 m/s. We can calculate the displacement during this time using the equation d = vt, where v is the velocity and t is the time.

By adding the displacements from each stage, we can find the total displacement of the car, which is the answer to the question.

Answer:

A spectrometer

Explanation: