A 1.00 x 109 kg object is raised vertically at a
constant velocity of 4.00 m/s by a crane. What
is the power output of the crane?

Answer:

the required current is 0.12 A

Option b) 0.12 A is the correct answer

Explanation:

Given the data in the question,

to determine the current needed in the wire, we use the following relation;

B = μ₀n[tex]I[/tex]

[tex]I[/tex] = B / μ₀n

where μ₀ is the magnetic constant   ( 4π × 10⁻⁷ T.m/A )

n is the number of turns ( 1 / 1mm[tex]\frac{10^{-3}m}{1 mm}[/tex]) = 1000 m⁻¹

B is magnitude ( 150μT ( [tex]\frac{10^{-6}m}{1uT}[/tex]) )

so we substitute

[tex]I[/tex] =  [ 150μT ( [tex]\frac{10^{-6}m}{1uT}[/tex]) ] / [ ( 4π × 10⁻⁷ T.m/A ) × 1000 m⁻¹ ]

[tex]I[/tex] =  [ 0.00015 ] / [ 0.00125 ]

[tex]I[/tex] = 0.12 A

Therefore, the required current is 0.12 A

Option b) 0.12 A is the correct answer

Answer:

# ICSE board exam 2021

Explanation:

# ICSE board exam 2021

# ICSE board exam 2021

# ICSE board exam 2021

Answer:

Positive, Negative

Explanation:

The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.

Answer:

a) [tex]\vec{d} =B-A[/tex]

b) [tex]I=\frac{B*v*d*\sin 90 \textdegree}{R}[/tex]

c) [tex]v \approx 0.6m/s[/tex]

Explanation:

From the question we are told that

Magnetic field strength [tex]B=3.6T[/tex]

Distance traveled [tex]d=7m[/tex]

Mass [tex]m=4.7 kg[/tex]

Resistance [tex]r=8.2 ohm[/tex]

Gravitational acceleration [tex]g=9.8m/s^2[/tex]

[tex]\theta =90 \textdegree[/tex] Because of perpendicularity

a)

Generally the direction of the current will be given as

[tex]\vec{d} =B-A[/tex]

Because it opposes increases of magnetic flux

b)

Generally the equation for induced EMF [tex]E[/tex] is mathematically given as

[tex]E=B*v*d*\sin \theta[/tex]

[tex]E=B*v*d*\sin 90 \textdegree[/tex]

Generally the equation for induced current [tex]I[/tex] is mathematically given as

[tex]I=E/R[/tex]

[tex]I=\frac{B*v*d*\sin 90 \textdegree}{R}[/tex]

c)

Generally the the equation for force F at terminal speed is mathematically given as

[tex]F=mg[/tex]

[tex]mg=B*I*d*\sin 90 \textdegree[/tex]

[tex]mg=B*(\frac{B * v * d }{R}) *d[/tex]

[tex]v=\frac{m*g*R}{B^2*D^2}[/tex]

[tex]v=\frac{4.7*9.8*8.2}{3.6^2*7^2}[/tex]

[tex]v=0.59475m/s[/tex]

[tex]v \approx 0.6m/s[/tex]

Answer:

60mph=26.8224meters per second

Explanation:

Explanation:

[tex]let \: coefficient \: of \: friction \: be \: \gamma \\ from \: third \: law \: of \: solid \: friction \\friction \: force = \gamma \times normal \: reaction \\ F = \gamma R \\ but \: R = mg = 240N \\ \gamma = \frac{ F}{R} \\ \gamma = \frac{12}{240} \\ \gamma = \frac{1}{20} [/tex]

Answer:

boiling water - When water heats, the heat from the burner is transferred to the container, heating the water at the bottom. This hot water rises, creating a circular motion, as colder water descends to replace it.

Answer:

The speed of the plank relative to the ice is:

[tex]v_{p}=-0.33\: m/s[/tex]

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

[tex]m_{g}v_{g}+m_{p}v_{p}=0[/tex] (1)

Where:

Now, as we have relative velocities, we have:

[tex]v_{g/b}=v_{g}-v_{p}=1.55 \: m/s[/tex] (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

[tex]45v_{g}+168v_{p}=0[/tex]

[tex]v_{g}-v_{p}=1.55[/tex]

[tex]v_{p}=-0.33\: m/s[/tex]

I hope it helps you!      

Answer:

Work done gets doubled.

Explanation:

The work done by a force is given by :

W = Fd

Where

F is force and d is distance move

If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

At the time when force is doubled and distance moved remain the same so Work done gets doubled.

The following information should be considered:

W = Fd

Where

F is force and d is distance move

In the case when the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

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Answer:

they employ front and Back defenses

Explanation:

there are the most positions in these types

Answer:

In my opinion, ''A" would be the longest wavelength.

Solve the differential equation:

dV/dt = k V   →   1/V dV/dt = k

→   d/dt [ln(V)] = k

→   ln(V) = k t + C

→   V (t )= exp(k t + C ) = C exp(k t ) = C e ᵏᵗ

At t = 0, the glacier has volume 400 km³ of ice, so

V (0) = 400   →   C e⁰ = C = 400

Find when the glacier's volume is 300 km³:

V (t ) = 400 e ᵏᵗ = 300   →   e ᵏᵗ = 3/4

→   k t = ln(3/4)

→   t = 1/k ln(3/4)

At this time, the volume is decreasing at a rate of 15 km³/yr, so

V ' (t ) = C k e ᵏᵗ   →   V ' (1/k ln(3/4)) = 400 k exp(k × 1/k ln(3/4)) = -15

→   3/4 k = -3/80

→   k = -1/20

Then the volume V (t ) of the glacier at time t is

V (t ) = 400 exp(-1/20 t )

The volume in terms of time will be "V(t) = 400 exp (-[tex]\frac{1}{20}[/tex] t)".

According to the question,

Glacier's volume, V(0) = 400 km³

→     [tex]\frac{dV}{dt}[/tex] = kV

   [tex]\frac{1}{V}[/tex] [tex]\frac{dV}{dt}[/tex] = k

   ln(V) = kt + C

Now,

   V (t) = exp (kt + C)

           = C exp (kt)

           = C[tex]e^{kt}[/tex]

When volume of glacier be "300 km³",

V(t) = 400 [tex]e^{kt}[/tex] = 300

[tex]e^{kt}[/tex] = [tex]\frac{3}{4}[/tex]

By taking log,

 kt = ln([tex]\frac{3}{4}[/tex])

   t = [tex]\frac{1}{k}[/tex] ln([tex]\frac{3}{4}[/tex])

When volume decrease at 15 km³/yr, then

→ V'(t) = Ck[tex]e^{kt}[/tex]

          = 400 k exp (k × [tex]\frac{1}{k}[/tex] kn ([tex]\frac{3}{4}[/tex]))

          = -15

Now,

[tex]\frac{3}{4}[/tex] k = - [tex]\frac{3}{80}[/tex]

By applying cross-multiplication,

  k = - [tex]\frac{1}{20}[/tex]

Thus the response above is correct.

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Answer:

AYE

Explanation:

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   [tex]\frac{m}{m+ \frac{I}{r^2} }[/tex]

          v² = 2gh    [tex]\frac{1}{1 + \frac{I}{m r^2} }[/tex]

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = [tex]g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }[/tex]

let's use trigonometry

         sin θ = h / L

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    [tex]\frac{1}{1 + \frac{mr^2 }{m r^2 } }[/tex]1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  [tex]\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }[/tex] = g sin θ   [tex]\frac{1}{1+ \frac{1}{2} }[/tex]

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   [tex]\frac{1}{1 + \frac{2}{3} }[/tex]

     a₃ = g sin θ [tex]\frac{3}{5}[/tex]

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  [tex]\frac{1 }{1 + \frac{2}{5} }[/tex]

     a₄ = g sin θ  [tex]\frac{5}{7}[/tex]

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      [tex]\frac{105}{210}[/tex]

b) a₂ = g sinθ ⅔ = g sin θ     [tex]\frac{140}{210}[/tex]

c) a₃ = g sin θ [tex]\frac{3}{5}[/tex]= g sin θ       [tex]\frac{126}{210}[/tex]

d) a₄ = g sin θ [tex]\frac{5}{7}[/tex] = g sin θ      [tex]\frac{150}{210}[/tex]

the order of acceleration from lower to higher is

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

The sun is made up mostly of  plasma material. Option D

Plasma is considered the fourth state of matter, distinct from gases, liquids, and solids. It is a highly ionized gas consisting of charged particles, including electrons and ions. The sun is primarily composed of plasma due to the extremely high temperatures and intense energy present in its core.

The sun's core has a temperature of about 15 million degrees Celsius (27 million degrees Fahrenheit), which is hot enough to strip electrons from atoms, creating a plasma state. The high temperature and pressure cause hydrogen atoms to undergo nuclear fusion, resulting in the release of vast amounts of energy in the form of light and heat.

This fusion process involves the conversion of hydrogen nuclei (protons) into helium nuclei, releasing enormous energy in the process. The intense heat and pressure within the sun's core sustain this fusion reaction, powering the sun and providing the energy that is radiated out into space as sunlight.

Plasma is an electrically conductive state of matter, meaning it can carry electric currents. The sun's plasma exhibits complex dynamics, including the generation of magnetic fields, solar flares, and coronal mass ejections.

In summary, the sun is primarily composed of plasma material due to the extreme temperatures and intense energy within its core. The plasma state allows for nuclear fusion to occur, releasing the immense energy that sustains the sun's radiative output.

Option D

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The amount of heat absorbed by the water will be 8368 J.

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.

It is given by the formula as ;

[tex]\rm Q= mcdt[/tex]

The given data in the problem is;

Equilibrium temperature = 30°C.

mass of water  = 50 g  ,

Temperature change = 70ºC

The specific heat of water =4.184 J//g °C

The amount of heat absorbed by the water is;

[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]

Hence, the amount of heat absorbed by the water will be 8368 J.

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Answer:

longer for less massive stars.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

Some of the examples of stars are Canopus, Sun (closest to the Earth), Betelgeus, Antares, Vega etc.

Generally, the time taken for the collapse of an interstellar cloud fragment to the period (time) when a main-sequence star is given birth to, is usually longer for less massive stars.

This ultimately implies that, stars that are not so massive or big in size are transformed from interstellar cloud fragment to a main-sequence star is lesser.

Time taken from, when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star is longer for less massive stars.

The process of Interstellar cloud fragment results the development into the stars.

The formation of stars is takes place within the highly dense concentration of the interstellar gases and the dust. These interstellar gases and the dust forms the molecular or interstellar clouds.

Here, in these molecular clouds the gases are present like helium, CO, hydrogen in major amount. Now this clouds started to collapse due to their own weight and density, and the formation of star is started.

Here, the H₂ atoms, which are fused forms helium atoms, to their core to form the main-sequence star.

In general, the time it takes from when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star. The time required to form the main-sequence star is lesser than the star which has big size.

Hence, the time taken from, when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star is longer for less massive stars.

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Answer:

110N north west

Explanation:

.........

If three forces F1 = 20N, 300NE, F2 = 50N along W, F3 = 40N 500NW act on a body. The resultant in magnitude and direction  is 110N north west

Force is a  quantitative parameter and it is an interaction between two physical bodies such as an object and its environment, there are different types of forces in nature.

it can be  defined as pushing or pulling of any object resulting from the object’s interaction or movement, without force the objects can not move

If an object in moving state the it will be either static or motion, the position of the object will only be changed if it is pushed or pulled and The external push or pull upon the object called as Force.

The contact force types  are  Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force

Non-Contact forces are occur from a distance  such as Electromagnetic Force, Gravitational Force, Nuclear Force

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Answer:

by radiation

by heating

electrically

mechanically

Answer:

Z  = (R^2 + X^2)^1/2

At the resonant frequency, the quantity X equal zero and

w L - 1 / (w * C) equals zero where w is omega the angular frequency

Answer:

-33750 N

Explanation:

Use coulomb's law: [tex]\frac{k(q1)(q2)}{r^{2} } = \frac{(9x10^{9})(-5x10^{-6})(3x10^{-6})}{0.002^{2} } = -33750 N[/tex]

Answer:

fffffgggggggggggggghhh

Answer:

A) q_o = 0.001 C

B) I = 0.001•e^(-t)

C) V_c = 1000e^(-t)

D) V_r = 1000e^(-t)

E) P = e^(-2t) watts

Explanation:

A) We are given;

Initial stored energy; U_o = 0.5 J

Capacitance; C = 1.0μF = 1 × 10^(-6) F

To find the charge, we will use the formula for energy in capacitors which is given by;

U = q²/2C

Thus, since we are dealing with initial energy, U is U_o and q is q_o

Making q the subject, we have;

q_o = √2CU_o

q_o = √(2 × 1 × 10^(-6) × 0.5)

q_o = 0.001 C

B) The charge as a function of time is expressed as;

q = q_o•e^(-t/RC)

Now the current is gotten by differentiating the charge function. Thus;

I = (q_o/RC)•e^(-t/RC)

Where;

R is Resistance = 1.0MΩ = 1 × 10^(6) Ω

C is capacitance = 1 × 10^(-6) F

(q_o/RC) is the initial current = 0.001/(1 × 10^(6) × 1 × 10^(-6))

(q_o/RC) = 0.001 A

Thus;

I = 0.001•e^(-t/(1 × 10^(6) × 1 × 10^(-6)))

I = 0.001•e^(-t)

C) Formula for potential difference across the capacitor is;

V_c = IR

I = 0.001•e^(-t)

R = 1 × 10^(6) Ω

Thus;

V_c = 1 × 10^(6) × 0.001•e^(-t)

V_c = 1000e^(-t)

D) Potential difference across the resistor will be the same as that across the capacitor because the resistor is connected in parallel to the capacitor.

Thus;

V_r = V_c = 1000e^(-t)

E) rate at which thermal energy is produced is basically the power.

Thus;

P = (V_r)²/R

P = (1000²e^(-2t))/1 × 10^(6)

P = e^(-2t) watts

Answer:

8.08 J/g °C

Explanation:

Q=m*Cp*ΔT-->

Cp=Q/(m*ΔT) -->

Cp=1600/[18*(31-20)]-->

Cp=8.08 J/g °C

Answer:

Displacement and Velocity

Answer:

Aluminum board will move with a higher velocity

Explanation:

The velocity of the block will be higher when the impulse imparted by the bullet is higher.

In case of bullet bouncing off, the impulse imparted on the aluminum board is high and hence, it will move with a high velocity as compared to that of the clay board.

Answer:

The scientist finds an equation that predicts how long the rainstrom will last.

Explanation:

I just answered it :)

Answer:

atoms

Explanation:

Hormones are derived from amino acids or lipids. Amine hormones originate from the amino acids tryptophan or tyrosine. Larger amino acid hormones include peptides and protein hormones. Steroid hormones are derived from cholesterol.

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

[tex]m'=9.00 \ g[/tex]

or,

    [tex]=0.009 \ kg[/tex]

mass of wooden block,

[tex]m=1.20 \ kg[/tex]

speed,

[tex]s=0.390 \ m[/tex]

Coefficient of kinetic friction,

[tex]\mu=0.20[/tex]

As we know,

The Kinematic equation is:

⇒  [tex]v^2=u^2+2as[/tex]

then,

Initial velocity will be:

⇒  [tex]u=v^2-2as[/tex]

        [tex]=v^2-2 \mu gs[/tex]

On substituting the given values, we get

⇒  [tex]u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}[/tex]

       [tex]=\sqrt{-1.5288}[/tex]

       [tex]=1.23 \ m/s[/tex]

As we know,

The conservation of momentum is:

⇒  [tex]mu=m'u'[/tex]

or,

⇒ Initial speed, [tex]u'=\frac{mu}{m'}[/tex]

On substituting the values, we get

⇒                            [tex]=\frac{1.20\times 1.23}{0.009}[/tex]

⇒                            [tex]=\frac{1.476}{0.009}[/tex]

⇒                            [tex]=164 \ m/s[/tex]