What resistance should be added in series with a 3.0-H inductor to complete an LR circuit with a time constant of 4.0 ms? A)0.75 k Ω Β) 12 Ω C) 0.75 Ω D) 2.5 Ω

Answer:

given us,

mass= 4×9.8gm m(9.8) formula

= 39.2

final velocity (v)= 60m/s

initial velocity (u)= 20m/s

time(t)= 10s

acleration(a)=?

now,

accleration(a)= v-u/t=60- 20/10

=40/10

=4m/s

:. the acceleration is 4 m/s

Explanation:

first we have to calculate mass and we can use acceleration formula

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.

The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.

A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.

The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.

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Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine,  

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

h = 0.24 meters

or

h = 24 cm

Answer:

The total torque is 27.5 Nm

Explanation:

Given;

5.0-N force directed upward is applied 4.0 m to the left of the pivot,

5.0-N force directed upward is applied 1.5 m to the right of the pivot,

Taking the moment about the pivot, the total torque is given by;

τ = Fr

where;

F is the appllied force

r is the radius of the force arm

τ = (5 N x 4 m) + (5 N x 1.5 m)

τ = 27.5 Nm

Therefore, the total torque is 27.5 Nm

Answer:

Silver

Explanation:

Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.

In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂

Answer:

I'm pretty sure its A

Explanation:

because its a reflection- Hope you get a good grade!

Answer:

I = 1.05x10⁻³ A

Explanation:

By definition, an electric current is the rate of charge flow at a given time:

[tex] I = \frac{q}{t} [/tex]

Where:

q: is the electrons charge = 1.602x10⁻¹⁹ C

t: is the time

In a circular motion, the time is given by:

[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]

Where:

ω: is the angular speed = v/r

v: is the speed = 2.19x10⁶ m/s

r: is the radius = 5.29x10⁻¹¹ m

[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]

Now, the effective current is:

[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]  

Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.

I hope it helps you!                                

Answer:abc defg hijk lmnop qrs tuv wx y and z

Explanation: now i know my abc's

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

From the question we have

[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]

We have the final answer as

Hope this helps you

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

Answer:

9.8m/s^2 down  (option C)

Explanation:

The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.

Answer:

Speed = 100 km/h

Explanation:

Given:

Speedometer reading = 100 kph for one lap

Assume;

Time taken to complete one lap = 1 hour

Computation:

Speed =  Distance / Time

Speed =  100 / 1

Speed = 100 km/h

Answer:

0.546 [tex]\hat k[/tex]

Explanation:

From the given information:

The force on a given current-carrying conductor is:

[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]

where the length usually in negative (x) direction can be computed as

[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]

[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]

[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]

where;

current I = 7.0 A

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]

F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]

F = 0.546 [tex]\hat k[/tex]

Answer:

h = 51.75 m

nearest answer is:

53.4 m

Explanation:

First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance = 39 m

V₀ₓ = Horizontal Initial Velocity = 12 m/s

t = time = ?

Therefore,

39 m = (12 m/s)t

t = 39 m/12 m/s

t = 3.25 s

Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:

h = V₀y t + (1/2)gt²

where,

h = height of cannon = ?

V₀y = initial vertical velocity = 0 m/s

g = 9.8 m/s²

Therefore,

h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²

h = 51.75 m

Explanation:

U = 40m/s

V = 30m/s

T = 5 sec

A = ?

[tex]a = \frac{u - v}{t}[/tex]

[tex]a = \frac{40 - 30}{5}[/tex]

[tex]a = \frac{10}{5}[/tex]

[tex]a = 2[/tex]

since it's decreasing in speed, The acceleration will be " - 2.0ms^-2 " or " - 2.0m/s^2 "

If you liked this answer, feel free to follow me for more!

Btw don't mind me answering twice. I want the free points and maybe another brainliest? lol.

Answer:

more than 51.45 N

__________________________________________________________

We are given:

Mass of the television = 15 kg

Coefficient of Static friction = 0.35

Minimum force required to move the television:

Normal Force:

We know that the normal force is equal and opposite to the Weight of the television

Weight of the television = Mg            

[where m is the mass and g is the acceleration due to gravity]

Weight = 15 * 9.8

Weight = 147 N

Force of Friction:

We are given the coefficient of Friction = 0.35

We know that coefficient of Friction = Force of friction / Normal Force

replacing the variables

0.35 = Force of Friction / 147

Force of Friction = 147 * 0.35                     [multiplying both sides by 147]

Force of Friction = 51.45 N

Since a force of 51.45 N is will be applied opposite to the direction of application of Force, the television will only move when we apply more force than 51.45 N

Answer:

it is C

Explanation:

Answer:

Nope

Explanation:

Answer:

0.967 m/s

1855 N

[tex]46.375\ \text{MPa}[/tex]

Explanation:

v = Final velocity

u = Initial velocity = 0

s = Displacement = 4.77 cm

g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

From the kinematic equations

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]

The velocity of the object at the moment of impact is 0.967 m/s

Now

[tex]\Delta v[/tex] = Change in velocity = 1 m/s

t = Time taken = 20 ms

m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]

[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]

The force associated with a single step of the person is 1855 N

A = Area = [tex]0.4\ \text{cm}^2[/tex]

Stress is given by

[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]

The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]

The speed of object during falling is 0.967 m/s.

(A)  The magnitude of force  associated with this single step for a person is 1855 N.

(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

Given data:

The height of fall is, h = 4.77 cm = 0.0477 m.

The magnitude of downward velocity is, v' = 1.0 m/s.

The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].

The mass of person is, m = 74.2 kg.

The magnitude of force is, F' = 1880 N.

The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].

The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,

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

Solving as,

[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]

Thus, the speed of object during falling is 0.967 m/s.

(A)

Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.

Apply the expression of average force as,

[tex]F =\dfrac{m'v'}{t}[/tex]

Solving as,

[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]

Thus, the magnitude of force  associated with this single step for a person is 1855 N.

(B)

The expression for the stress is given as,

[tex]\sigma = \dfrac{F'}{A}[/tex]

Solving as,

[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]

Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

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Answer:

Explanation:

pV = nrT

n = PV/RT

n = (100*10^3)(.01)/(300*0.082057)

n = 40.62 moles

Answer:

this is a projectile

Answer:

D. projectile

Explanation:

Since it's dropped from a rest that means that it's velocity at the beginning is 0.

131.2 J and The last one on number 8

I gave the same answer and it passed.

7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.

8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Given:

A moving object is rolling on a surface that is 5 m off the ground,

The speed of the object, v = 4 m/s,

The mass of the object, m = 3.2 kg,

Calculate the kinetic energy after 10 meters as shown below,

KE = 1/2 × 4² × 3.2

KE = 25.6 J

Calculate the potential energy as shown below,

PE = 3.2 × 9.8 × 5

PE = 156.8 J

Thus, total energy = KE + PE

The total energy = 25.6 + 156.8

The total energy = 182.4 J

8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.

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Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

Answer:

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]

Answer:

lucky mauld mauldgomary was an british poet...

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

Answer:Acceleration implies any change in the velocity of the object with respect to time. Velocity is nothing but the rate of change of displacement. On the other hand, acceleration is the rate of change of velocity with respect to time.

Explanation: