g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s

Answers

Answer 1

Answer:

The speed of m2 just before it hits the ground is 2.1 m/s

Explanation:

mass on the ground m1 = 30 kg

mass oat rest at the above the ground m2 = 35 kg

height of m2 above the ground =2.9 m

Let the tension on the string be taken as T

for the mass m2 to reach the ground, its force equation is given as

[tex]m_{2} g - T = m_{2}a[/tex]    ....equ 1

where g is acceleration due to gravity = 9.81 m/s^2

and a is the acceleration with which it moves down

For mass m1 to move up, its force equation is

[tex]T - m_{1} g = m_{1} a[/tex]

[tex]T = m_{1}a + m_{1}g[/tex]

[tex]T = m_{1}(a + g)[/tex]    ....equ 2

substituting T in equ 1, we have

[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]

imputing values, we have

 [tex](35*9.81) - 30(a+9.81) = 35a[/tex]

 [tex]343.35 - 30a-294.3 = 35a[/tex]

[tex]343.35 -294.3 = 35a+ 30a[/tex]

[tex]49.05 = 65a[/tex]

a = 49.05/65 = 0.755 m/s^2

The initial velocity of mass m2 = u = 0

acceleration of mass m2 = a = 0.755 m/s^2

distance to the ground = d = 2.9 m

final velocity = v = ?

using Newton's equation of motion

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

substituting values, we have

[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]

[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]

v = 2.1 m/s


Related Questions

g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride

Answers

Answer:

2.52 rpm

Explanation:

given that

diameter of the wheel, d = 25 m

Mass of the person, m = 75 kg

Weight experienced, N = 810 N

Since diameter is 25, radius then is 25/2 = 12.5 m

We all know that,

v = rw

Also, the passengers weight is equal to the centripetal acceleration, and thus

mg = mv²/r

Substitute for v, we have

mg = m/r * (rw)²

mg = mr²w²/r

g = rw²

If we make w the subject of formula, we have

w² = g/r

w = √(g/r)

mg = 810

75 * g = 810

g = 810 / 75

g = 1.08 m/s²

w = √(g/r)

w = √(1.08 / 12.5)

w = √0.0864

w = 0.294 rad/s

Since the question asked us in rpm, we convert to rpm

0.294 * (60 / 2π)

2.52 revolution per minute.

What portion of the difference in the angular speed before and after you increased the mass can be accounted for by frictional losses

Answers

Answer:

As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.

Explanation:

friction can be defined as resistance in motion of bodies in relative to one another

momentum is the product of mass and velocity

torque is the time rate of change in momentum

τ = [tex]\frac{dL}{dt}[/tex]

where L = Iω = mvr

I = moment of inertia

ω=  angular frequency

if there is no external force(torque) acting on the system, then

[tex]\frac{dL}{dt}[/tex] = 0

dL = 0 = constant

moment of inertia I depends on the distribution of mass on the axis of rotation.

as the mass increases, the angular momentum(L) increases

angular frequency, ω, remains constant

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?

Answers

Answer:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

            [tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

   

Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?

Answers

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

       

When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answers

Answer:

The  internal resistance is  [tex]r = 0.5 \ \Omega[/tex]

Explanation:

From the question we are told that the resistance of

   The  resistance of the resistor is  [tex]R = 20.0\ \Omega[/tex]

    The  voltage is [tex]V = 12.0 \ V[/tex]

     The magnitude of the voltage fall is  [tex]e = 0.300\ V[/tex]

Generally the current flowing through the terminal due to the voltage of the battery  is  mathematically represented as

        [tex]I = \frac{V}{R}[/tex]

substituting values

        [tex]I = \frac{12.0 }{20 }[/tex]

       [tex]I = 0.6 \ A[/tex]

The internal resistance of the battery is mathematically represented as

      [tex]r = \frac{e}{I}[/tex]

substituting values

     [tex]r = \frac{0.300}{ 0.6 }[/tex]

    [tex]r = 0.5 \ \Omega[/tex]

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

(V/R)r = V'............. Equation 1

Where:

V = Voltage across the terminal of the batteryR = Resistance connected across the batteryr = internal resistance of the batteryV' = voltage drop of the battery.

Make r the subject of the equation

r = V'R/V............ Equation 2

From the question,

Given:

V = 12 VR = 20 ohmsV' = 0.3 V

Substitute these values into equation 2

r = (0.3×20)/12r = 6/12r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

Learn more about internal resistance here: https://brainly.com/question/14883923

Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed

Answers

Answer:

The number of turns of the wire needed is 243 turns

Explanation:

Given;

self inductance of the solenoid, L = 1.8 mH

cross sectional area of the inductor, A = 1.6 x 10⁻³ m²

length of the inductor, l = 0.066 m

The self inductance of long solenoid is given by;

L = μ₀n²Al

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

n is number of turns per length

A is the area of the solenoid

l is length of the solenoid

[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]

The number of turns is given by;

N = nL

N = (3682.74)(0.066)

N = 243 turns

Therefore, the number of turns of the wire needed is 243 turns

1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?

Answers

Answer:

Explanation:

let the coil of length l and breathe b entering the magnetic field B with speed v.

So, the magnetic flux through the coil is

Ф = B(l×b)

length × breathe = area

Ф = BA

dФ = BdA

therefore induced emf is given as

ε = [tex]Bl(\frac{db}{dt})[/tex]

note: [tex]\frac{db}{dt} = v[/tex]

ε[tex]= Blv[/tex]

attached is the diagram for the solution

A trash compactor can compress its contents to 0.350 times their original volume and 4 times denser than their original density. Neglecting the mass of air expelled, what factor is the old density of the rubbish

Answers

Answer:

2.8

Explanation:

Using p = m/v; (old density)

p' = m/v (new density)

=m/0.350 V

p'/p = (m/0.350V)/(m/v) = 1/0.350 = 2.86

which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.

Answers

Answer:

D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity

Explanation:

Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.


I need help with this question?

Answers

Answer:

You got it right, didn't you?

c) vector C

Explanation:

opposite to vector V

Answer:

A, vector B

Explanation:

A negative vector is a vector which points in the opposite direction, even though it’s in a different quadranot it’s still the opposite direction.

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron

Answers

Answer:

The speed of the electron is 6.79 x 10⁵ m/s

The radius of the circular path is 1.357 x 10⁻⁵ m

Explanation:

Given;

magnetic field, B = 0.285 T

energy of electron, E = 2.10 x 10⁻¹⁹ J

The kinetic energy of the electron is calculated as;

[tex]K.E = \frac{1}{2} m_eV^2[/tex]

Where;

[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg

V is the speed of the electron

[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]

The radius of the circular path is given by;

[tex]R = \frac{M_eV}{qB}[/tex]

where;

q is the charge of the electron = 1.6 x 10⁻¹⁹ C

[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]

You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6 cm distributed uniformly along its 33 cm length.
A. How much current is needed?
_____________ A
B. What is the magnitude of the magnetic field inside the solenoid?
________________T
C. What is the energy density (energy/volume) inside the solenoid?
________________ kJ/m^3

Answers

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

[tex]L= \frac{\mu_o N^2 A}{l}[/tex]

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

[tex]L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H[/tex]

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

[tex]E = \frac{1}{2} LI^2\\\\[/tex]

Where;

I is the current in the solenoid

[tex]E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A[/tex]

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

[tex]U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3[/tex]

Suppose I am viewing light through a camera lens (i.e. a circular aperture). If I want a wider field of view I should _____ the diameter of the lens.

Answers

Answer:

Increase

Explanation:

Because For a given focal length, a lens with a larger front element will generally be faster. That is, it'll have a larger maximum aperture, allowing a shorter exposure time, But a larger aperture requires larger elements to maintain the same angle of view

A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.

Answers

Answer:

As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.

1. weight

2. gravitational potential energy to kinetic energy.

Explanation:

As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.

work done by toboggan = weight × distance

W = mg and the distance is down the icy slope

By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.

Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)

Suppose a 225 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m. How high can it coast up the hill, if you neglect friction in m?
a) m = 180 kg
b) v = 29 m/s
c) h = 32 m

Answers

Answer:

It can coast uphill 6.2m

Explanation:

See attached file pls

A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and the pump efficiency is 85%, the power required by the pump is most nearly:

Answers

Answer:

The power required by the pump is nearly 230.588 kW

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of [tex]P_{p}[/tex]

where [tex]P_{p}[/tex] is the power of the pump needed

85% = 0.85

196000 = 0.85[tex]P_{p}[/tex]

[tex]P_{p}[/tex] = 196000/0.85 = 230588.24 W

Pump power = 230.588 kW

Wind gusts create ripples on the ocean that have a wavelength of 5.00 cm and propagate at 2.00 m/s. What is their frequency

Answers

Answer:

f = 40Hz

Explanation:

v=f x wavelength

f =v / wavelength

f = 2/5 x 10-²= 40 Hz

f = 40Hz

f = 40Hz

What is frequency?

In physics, the term frequency refers to the number of waves that pass a fixed point in unit time.

It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.

Wavelength (λ) - The wavelength of light is defined as the distance between the crests or troughs of a wave motion.

The wave equation: v = fλ

As per question,

Wavelength = 5.00 cm

v = 2.00 m/s.

v=fλ

f =v / λ

f = 2/5 x 10⁻² = 40 Hz

f = 40Hz

Therefore,

The frequency is 40Hz.

Learn more about frequency here:https://brainly.com/question/14316711

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A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s

Answers

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.

Answers

Answer:

Explanation:

The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.

In compound microscopes, the distance between the two lenses is expressed as L = v0+ue

v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.

Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.

Given L = 18.0cm

Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm

1/f0 = 1/u0+1/v0

f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.

1/v0 = 1/7.4-1/12

1/v0 = 0.1351-0.0833

1/v0 = 0.0518

v0 = 1/0.2184

v0 = 19.31cm

Note that v0 = ue = 19.31cm

To get ve, we will use the lens formula 1/fe = 1/ue+1/ve

1/ve = 1/fe-1/ue

Given ue = 19.31cm and fe = -7.00cm

1/ve = -1/7.0-1/19.31

1/ve = -0.1429-0.0518

1/ve = -0.1947

ve = 1/-0.1947

ve = -5.14cm

Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens

b) Magnification of the final image M = ve/ue

M = 5.14/19.31

M = 0.27

Magnification of the final image is 0.27

a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's magnetic field, the ion will be delfected:

Answers

Answer:

East direction

Explanation:

Given that

Charge on the particle is positive.

Moving towards the center of earth .

We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.

Therefore due to the earth magnetic fields , the positive ions will deflect  towards  East direction.

Thus the answer will be East direction.

If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B

Answers

Answer:

0.9m/s²

Explanation:

See attached files

A Buchner funnel uses _______ when separating a(n) _______ by filtration.

Answers

Explanation:

A Buchner funnel uses perforatet glass plate when separating a(n) solide from liquid by filtration.

[tex]hope \: this \: helps[/tex]

A charged particle moves into a region of uniform magnetic field B (pointing out of the page), goes through half a circle, and exits the region. The particle is either a proton or an electron. It spends 130 ns in the region. (a) What is the magnitude of B

Answers

The figure is missing, so i have attached it

Answer:

Magnitude of B = 0.252 T

Explanation:

From the image, considering the point at which it enters the field-filled region, the velocity vector is pointing downwards. The field points out of the page so that; (v→) × (B→) points leftward, points leftward which indeed seems to be the direction it is pushed. Therefore q > 0 and thus it's a proton.

The equation for the period since it goes through half circle is;

T = 2t = 2πm/(e|B|)

Where;

m is mass of proton = 1.67 × 10^(-27) kg

e is electron charge = 1.60 x 10^(-19) Coulombs.

|B| is magnitude of magnetic field

t = 130 ns = 130 × 10^(-9) s

Making |B| the subject, we have;

|B| = πm/et

Thus, plugging in all relevant values, we have;

|B| = π(1.67 × 10^(-27))/(1.60 x 10^(-19) × 130 × 10^(-9)) = 0.252 T

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that places them in series across normal household voltage. If each one is an incandescent bulb of fixed resistance, which statement about these bulbs is correct?

Answers

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is [tex]P_1 = 100 \ W[/tex]

      The power rating of the second bulb is  [tex]P_2 = 50 \ W[/tex]

     

Generally the power rating of the first bulb is mathematically represented as

      [tex]P_1 = V^2 R[/tex]

Where  [tex]V[/tex] is the normal household voltage which is constant for both bulbs

  So  

        [tex]R_1 = \frac{V^2}{P_1 }[/tex]

substituting values

        [tex]R_1 = \frac{V^2}{100}[/tex]

Thus the resistance of the second bulb would be evaluated as

       [tex]R_2 = \frac{V^2}{50}[/tex]

From the above calculation we see that

        [tex]R_2 > R_1[/tex]

This power rating of the first bulb can also be represented mathematically as  

        [tex]P_ 1 = I^2_1 R_1[/tex]

This power rating of the first bulb can also be represented mathematically as    

       [tex]P_ 2 = I^2_2 R_2[/tex]

Now given that they are connected in series which implies that the same current flow through them so

       [tex]I_1^2 = I_2^2[/tex]

This means  that

       [tex]P \ \alpha \ R[/tex]

So  when they are connected in series

     [tex]P_2 > P_1[/tex]

This means that the 50 W bulb glows more than the 100 \ W bulb

It takes 144 J of work to move 1.9 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates

Answers

Answer:

151.58 V

Explanation:

From the question,

The work done in a circuit in moving a charge is given as,

W = 1/2QV..................... Equation 1

Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.

make V the subject of the equation

V = 2W/Q.................. Equation 2

Given: W = 144 J. Q = 1.9 C

Substitute into equation 2

V = 2(144)/1.9

V = 151.58 V

You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?

Answers

Answer:

I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows

Explanation:

In thermodynamics, coolness can not be transferred, only heat can be transferred

Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.

Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field

Answers

Answer:

case 1 of physics is the answer

The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.

Answers

Answer:

27°

Explanation:

The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)

So theta = arcsin(0.45)

=27°

The angle between the wire and the magnetic field is 27°.

Calculation of the angle:

Since The magnetic force per meter on a wire is measured to be only 45 %

So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field

Therefore,

theta = arcsin(0.45)

=27°

Hence, The angle between the wire and the magnetic field is 27°.

Learn more about wire here: https://brainly.com/question/24733137

A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?

Answers

Answer:

Convection and Radiation mechanisms carry most of the heat

Explanation:

This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?

Answers

Answer:

31

Explanation:

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