A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top of it as shown. The block w2 is attached to avertical wall by a string 6m long. If the coefficient of friction between all surface is 0.25 and the system is in equilibrium find the magnitude of the horizontal force applied to the lower block

Answer:

Explanation:

       Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.

           #I AM ILLITERATE

Explanation:

bro I have no idea fam......

Answer:

well its about our thinking but i do believe in ghost a little

Answer:

there are 3 photos attached. so check

Explanation:

Answer:F = 255 N

Explanation:

It is given that,

Mass of the boy, m = 25 kg

Acceleration of the elevator,  

The elevator is accelerating in upward direction. The net force acting on the boy is given by :

g is the acceleration due to gravity

F = 255 N

The scale reading is 255 N as it begins to accelerate upward. hence, this is the required solution.

The question is incomplete. The complete question is :

A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.

Solution :

Given :

Length of the wire, L = 0.6 m

Mass of the wire length, m = 11 g

                                             = [tex]11 \times 10^{-3}[/tex] kg

Magnetic field , B = 0.4 T

Know we know that :

ILB = mg

or [tex]$I=\frac{mg}{BL}$[/tex]

 [tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]

 [tex]I=0.44963\ A[/tex]

 [tex]I = 449.63 \ mA[/tex]

Explanation:

I=2.5 Ampere ; V=100V ;t = 1 hour=60secs

We know Heat = VIt

H=100×2.5×60=15,000J

Answer:

Explanation:

150/5  = 30

30mph per 1 second

I assume friction is the only force acting on the book as it slides.

(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²

W ≈ -6.56 J

(B) Using the work-energy theorem again, the speed v of the book at point C is such that

-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v = 0.750 m/s

(C) Take the left side to be positive, then solve again for v.

0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v ≈ 1.60 m/s

Answer:

pendulum, quartz

Explanation:

We have that the  energy (in calories) is transferred to the water as heat,to the bowl and  the original temperature of the cylinder  is mathematically given as

Energy

Generally the equation for the   is mathematically given as

(a)

The heat transferred to the H20

Qw= CwMwdT+Lvms

Qw=((220g)(100°C-20.0T)+(539 caVg)(5.00 g)

Qw=20.3 kcal .

(b)

The heat transferred to the bowl is

Qb= CbmbdT

Q= (0.0923 cal/gC)(150g)(100°C-20.0°C)

Q= 1.11 kcal

(c)  

original temperature of the cylinder

-Qw- Qb = CcMc(T2-T1)

[tex]T1=\frac{Qw+Qm}{CcMc}+T2[/tex]

T1=873C

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Answer:

Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.

Hope this helps..

The frequency is 4,6E14 Hz.

What is the frequency?

Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.

Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.

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Complete Question

(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?

Answer:

[tex]Q=4.50 *10^7J[/tex]

Explanation:

From the question we are told that:

Time [tex]t=5hours[/tex]

Temperature rise [tex]dT= 65\textdegree[/tex]

Power [tex]P=2500.0 Watts[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=\frac{Q}{t}[/tex]

Therefore

[tex]Q=2500*5*360[/tex]

[tex]Q=4.50 *10^7J[/tex]

Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.

Here,

Mass of the car, m = 0.5 kg

Velocity of the car, v = 6 m/s

Radial friction between the car and the track, f = 7 N

The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.

So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.

So,

mv²/r > f

0.5 x 6²/r > 7

Therefore, the radius of the circular track,

r < 18/7

r < 2.57 m

Hence,

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

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Answer:

0.032 [tex]m/s^2[/tex]

Explanation:

Given :

Weight of the astronaut = 69 kg

Weight of the physics book = 28 kg

Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]

The force that is applied on the astronaut :

[tex]F=ma[/tex]

   [tex]$=69 \times 0.013$[/tex]

   = 0.897 N

Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N

Therefore, the acceleration of the physics book is given by :

[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]

[tex]$a = \frac{0.897}{28}$[/tex]

a = 0.032 [tex]m/s^2[/tex]

Hence, the acceleration of the physics book is  0.032 [tex]m/s^2[/tex].

Answer:

The acceleration of astronaut is 5.27 x 10^-3 m/s^2.

Explanation:

mass of astronaut, M = 69 kg

Mass of book, m = 28 kg

acceleration of book, a = 0.013 m/s^2

Let the acceleration of astronaut is A.

According to the Newton's third law, for every action there is an equal and opposite reaction.

So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.

M A = m a

69 x A = 28 x 0.013

A = 5.27 x 10^-3 m/s^2

Answer:

It tell you that the velocity is constant, what means that there's no acceleration

Answer:

C. Mass

Explanation:

Answer:

[tex]5/27[/tex]

Explanation:

wavelengths for Lyman series

[tex]\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}[/tex]

wavelengths for Balmer series

[tex]\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}[/tex]

[tex]\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27[/tex]

OAmalOHopeO

The ratio of longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is 5/27. The correct option is 3.

Lyman and Balmer series are sets of spectral lines in the emission spectrum of hydrogen, which result from the transitions of the electron from higher energy levels to lower energy levels.

The Lyman series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=1 energy level. These transitions release energy in the form of ultraviolet photons. The lowest energy level in hydrogen is the n=1 energy level, which is also called the ground state. Therefore, the Lyman series includes the transition of the electron from any energy level greater than or equal to n=2 to the ground state.

The Balmer series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=2 energy level. These transitions release energy in the form of visible photons. The lowest energy level in the Balmer series is the n=2 energy level. Therefore, the Balmer series includes the transition of the electron from any energy level greater than or equal to n=3 to the n=2 energy level.

Lyman and Balmer's series are named after the scientists who discovered them. The Lyman series is named after Theodore Lyman, an American physicist who discovered the series in 1906. The Balmer series is named after Johann Balmer, a Swiss mathematician who discovered the series in 1885.

Here in the Question,

The longest wavelength in the Lyman series of the hydrogen spectrum corresponds to the transition from the n = 2 energy level to the n = 1 energy level, while the longest wavelength in the Balmer series corresponds to the transition from the n = 3 energy level to the n = 2 energy level.

The wavelengths of these transitions can be calculated using the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

where λ is the wavelength of the photon emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron.

For the longest wavelength in the Lyman series, we have n1 = 2 and n2 = 1, so:

1/λ_lyman = R(1/2^2 - 1/1^2) = 3R/4

For the longest wavelength in the Balmer series, we have n1 = 3 and n2 = 2, so:

1/λ_balmer = R(1/3^2 - 1/2^2) = 5R/36

Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is:

λ_lyman/λ_balmer = (3R/4)/(5R/36) = 27/20

Simplifying this ratio gives:

λ_lyman/λ_balmer = 27/20

Multiplying both the numerator and denominator by 1/3R, we get:

λ_lyman/λ_balmer = (1/2)/(1/3) = 3/2

Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is 3:2, or 3/5 in fractional form. Simplifying this ratio gives:

λ_lyman/λ_balmer = 5/3

Taking the reciprocal of both sides, we get:

λ_balmer/λ_lyman = 3/5

Therefore, the correct answer is (3) 5/27.

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Answer:

you will stc ohxoyxct txxtx xigigjjgjvvixiffjz,iffzikzfjvixii. hi h ohigiogooigoh

Explanation:

k jjvhvojvovvojivivivihvi hj

i. [tex]T = 36.8\:\text{N}[/tex]

ii. [tex]a = 2.45\:\text{m/s}^2[/tex]

iii. [tex]x = 1.23\:\text{m}[/tex]

Explanation:

Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]

Forces acting on m1:

[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]

Forces acting on m2:

[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]

Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us

[tex]m_2g - m_1g = m_2a + m_1a[/tex]

or

[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]

Solving for a,

[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]

[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]

We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).

[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]

Assuming that the two objects start from rest, the distance that they travel after one second is given by

[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]

Answer:

I hope this is correct!

Explanation:

a) work = (force)(displacement)

We know that d = 5m, now we just need to find the force

We need to calculate the acceleration first using a = v^2 - u^2 / 2d

final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m

a = 2.1^2 - 0^2 / 2(5)

  = 0.441 m/s^2

Now we can find force using f = ma

f = (14kg)(0.441m/s^2)

 = 6.174 newtons

Finally, you can calculate work now that you have force!

work = (6.174n)(5m)

       = 30.87 J (you may need to round sig figs!)

b) Work done by friction = (coefficient of kinetic friction)(mg)(v)

m = 14kg, g = 9.8, coefficient of friction = 0.2

force due to friction = (0.2)(14kg x 9.8)(2.1)

                                 = 57.624 J (again you may need to round to sig figs)

c) Work you perform = total work in direction of motion + work done by friction

total work = 30.87 J, work done by friction = 57.624 J

Work you perform = 30.84 + 57.624

                               = 88.464 J

d) Force you applied = work you performed / distance

work you performed = 88.464 / 5 m

                                   = 17.6929 N (again you may need to round to sig figs)

Hope this helps ya! Best of luck <3

Answer:

 law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

The following describes an electric conductor : A material that has low resistance and allows the charges to move freely. The correct option is D.

A conductor is a material or metal which allows the electrons to flow through it. In other words, a conductor allows the current to pass through them.

A battery also called as the voltage source, provides sufficient voltage or energy to excite electrons in the conductor.

Opposition offered to the flow of current is called as the resistance. The electrical element used in the circuit is the resistor.

So, an electric conductor is a material that has low resistance and allows the charges to move freely.

Thus, the correct option is D.

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Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

Answer:

1. True

2. False

3. True

4. True

5. True

Answer:

that is pure falsereeeeeeeee

Explanation:

Explanation:

Recall that

[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]

and

[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]

From Eqn(2), we can write

[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]

Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as

[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]

Squaring the above equation, we get

[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]

[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]

Use Eqn(3) on Eqn(4) and we will get the following:

[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]

This simplifies to

[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]

Rearranging this further, we get

[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]

Putting [tex]v_0^2[/tex] to the left side, we get

[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]

Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as

[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]

Answer:

use light of the same wavelength but decrease it's intensity

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]

Therefore,

[tex]$AD-BD = \frac{\lambda}{2}[/tex]

[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2 \times 0.4985$[/tex]

[tex]$\lambda = 0.99714 \ m$[/tex]

Thus the frequency is :

[tex]$f=\frac{v}{\lambda}$[/tex]

[tex]$f=\frac{340}{0.99714}$[/tex]

[tex]f=340.9744[/tex] Hz

Answer:

answer

Explanation:

it is the answer which was presented in the year