A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 N/m. The acceleration of gravity is 9.8 m/s2.
How far will the second spring compress in order to bring the mass to a stop?

Answer:

option B and C

Explanation:

Control rod are used to regulate the nuclear reactor.

When you insert control rod in the reactor it slows down the nuclear fission inside the reactor and the energy produced in the reactor will be less.

When you remove control road from the reactor the nuclear fission increase inside the reactor and the energy production is high.

Control rod consist of boron, boron absorb the neutron which help to control the nuclear fission.

Hence, the correct answer is option B and C

Answer:

a)P =14288.4 W

b)P = 19.16  horsepower

Explanation:

Given that

m= 108 kg

Initial velocity ,u= 0 m/s

Final velocity ,v= 19.6 m/s

t= 2.9 s

Lets take acceleration of Cheetah is a m/s²

We know that

v= u  + a t

19.6 = 0 + a x 2.9

a= 6.75 m/s²

Now force F

F= m a

F= 108 x 6.75 N

F= 729 N

Now the power P

P = F.v

P = 729 x 19.6 W

P =14288.4 W

We know that

1 W= 0.0013  horsepower

P = 19.16  horsepower

P =14288.4 W

Answer:0.25 kg-m/s

Explanation:

Given

mass of blob [tex]m=50 gm [/tex]

initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]

time of collision [tex]t=20 ms[/tex]

we know Impulse is equal to change in momentum

initial momentum [tex]P_i=mu[/tex]

[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]

Final momentum [tex]P_f=50\times 10^{-3}v[/tex]

[tex]P_f=0[/tex] as final velocity is zero

Impulse [tex]J=P_f-P_i[/tex]

[tex]J=0-(-0.25)[/tex]

[tex]J=0.25 kg-m/s[/tex]

The purpose of system is to make system more reliable and efficient for desired work completion

Explanation:

In most of the times, the system analysts operate in a dynamic environment where change is  the necessary process . A business application, a business firm or a computer system may be the required environment. In order To rebuild a system, the key elements must be considered which we need to examine are as follows:

1. Outputs and inputs.

2. Security and Control.

3. processor

4. Environment.

5. Feedback.

6. Boundaries and interface.

Answer:

gdsz

Explanation:

dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5

Final answer:

The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.

Explanation:

To calculate the moment of inertia of the wheel, we can use the principle of conservation of energy. The initial gravitational potential energy of the mass is equal to the final rotational kinetic energy of the wheel. This can be represented by the equation:

mg * h = 1/2 * Iω^2

Where m is the mass, g is the acceleration due to gravity, h is the distance the mass has descended, I is the moment of inertia of the wheel, and ω is the angular velocity of the wheel. Rearranging the equation:

I = 2mg * h / ω^2

Substituting the given values:

I = 2 * 6.32 kg * 9.8 m/s^2 * 2.5 m / (4.0 m/s)^2

I = 0.964 kg * m^2

Therefore, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.

Answer:

μk = 0.488

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the inclined plane

fk : kinetic Friction force: parallel to the inclined plane

Calculated of the W

W= m*g

W= 4.9 kg* 9.8 m/s² = 48.02 N

x-y weight components

Wx = Wsin θ = 48.02*sin27° = 21.8 N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 42.786 N

Calculated of the fk

fk = μk* FN=  μk*42.786 Equation (1)

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :

d = v₀*t+(1/2)*a*t² Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

t: time interval   (m/s)

a: acceleration ( m/s²)

Data:

d= 2.7 m

v₀ = 0

t= 5.4 s

We replace data in the formula (2)  

d = v₀*t+(1/2)*a*t²

2.7 = 0+(1/2)*a*( 5.4)²

2.7 = (14.58)*a

a = 2.7 / (14.58)

a= 0.185 m/s²

We apply the formula (1) to calculated μk:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-fk= m*a     , fk=μk*42.786 of the Equation (1)

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

μk = (20.89) / (42.786)

μk = 0.488

Answer:

number of molecules=  2.83 x 10^19

Explanation:

0.04 % means 0.04 L CO_2 in 100 L atmosphere

so for 2.9 L atmosphere CO2 vol = ( 0.04/100) x 2.9 = 0.00116

T = 27°C = 300 K , P = 1 atm ,

n = PV/RT = ( 1 x 0.00116) ÷ ( 0.08206x300) = 4.71×10^{-5}

number of molecules = 6.023 x 10^23 x 4.71×10^{-5}   = 2.83 x 10^19

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

Final answer:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.

Explanation:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.

Answer:

The final velocities of all the six balls will be same.

Explanation:

According to law of conservation of energy:

Gain in K.E = Loss in potential energy

   ½ mv^2 = mgh  

Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.

The six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.

During a downward motion of an object, the speed of the object increases as the object moves downwards and becomes maximum before the object hits the ground.

The equation for estimating the final velocity of the six balls is given as;

[tex]v_f = v_i + gt[/tex]

If air resistance is negligible, the six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.

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Answer: The temperature increases on the stratosphere with the altitude, given that absorption of the ultraviolet rays by the ozone.

Explanation: On the stratosphere, the water vapor and the umidity are almost nonexistents and, in view of the absorption of ultraviolet rays by the ozon, the temperature increases, reaching 35,6º Fahrenheit.  

The ozone is a unusual type of oxygen molecule. In the stratosphere, the ozone appears on a large scale and warms it up by the absorption of the ultraviolet rays energy.

The temperature in the stratosphere increases with altitude due to the absorption of UV radiation by ozone, which leads to exothermic reactions that generate heat, especially at higher altitudes where ozone concentration is greater.

The temperature increases with height in the stratosphere primarily because of the absorption of ultraviolet (UV) radiation by ozone. Ozone, a form of oxygen molecule, is very good at absorbing UV radiation from the Sun. When UV radiation is absorbed by ozone, it leads to exothermic chemical reactions which generate heat, thus raising the temperature of the surrounding atmosphere.

At higher altitudes in the stratosphere, the concentration of ozone is greater and, consequently, more UV radiation is absorbed, leading to a higher temperature in this region compared to the lower part of the stratosphere. It is this temperature inversion that distinguishes the stratosphere from the troposphere below, where the temperature normally decreases with altitude. The absorption of UV radiation and its conversion into heat makes the top of the stratosphere hotter as it is closer to the source of shortwave radiation, and the amount of absorbed shortwave radiation diminishes towards the bottom of the stratosphere.

Answer: [tex]4.8(10)^{-5} m[/tex]

Explanation:

We can solve this problem by the following equation:

[tex]\eta=\frac{F.h}{A \Delta x}[/tex]

Where:

[tex]\eta=3.5(10)^{10}Pa[/tex] is the shear modulus for brass

[tex]F=4.2(10)^{4}N[/tex] is the applied force

[tex]h=2.5 cm=0.025 m[/tex] is the height of the cube

[tex]A=h^{2}=(0.025 m)^{2}=625(10)^{-6} m^{2}[/tex] is the area of each surface of the cube

[tex]\Delta x[/tex] is the shear displacement

Finding [tex]\Delta x[/tex]:

[tex]\Delta x=\frac{F.h}{A \eta}[/tex]

[tex]\Delta x=\frac{(4.2(10)^{4}N)(0.025 m)}{(625(10)^{-6} m^{2})(3.5(10)^{10}Pa)}[/tex]

Finally:

[tex]\Delta x= 4.8(10)^{-5} m[/tex]

Answer:

Mass attached to the spring is 41.95 kg

Explanation:

We have given time period of the spring T = 2.1 sec

Let the mass attached is m

And spring constant is k

We know that time period is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1

Now if the mass is increased by 68.10 kg then time period become 3.4 sec

So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]

[tex]0.381=\frac{m}{m+68.10}[/tex]

[tex]m=41.95 kg[/tex]

So mass attached to the spring is 41.95 kg

Final answer:

To find the value of mass m, use the formula for the period of a mass-spring system.

Explanation:

In order to find the value of mass m, we can use the formula for the period of a mass-spring system:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the spring constant.

For the initial system with period 2.1 seconds, we have:

2.1 = 2π√(m/k)

For the system with mass increased by 6.810×10^1 kg and period 3.4 seconds, we have:

3.4 = 2π√((m + 6.810×10^1)/k)

Using these two equations, we can solve for the value of m.

Final answer:

The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.

Explanation:

To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.

The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).

For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:

W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)

Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.

W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)

W = 3.897×1010 J

Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.

Answer:

(ii) 1s2 2s2 2p6 3s2

Explanation:

Electron Affinity is the energy change that occur when an atom gains an electron.

                                  X₍₉₎ + e⁻  →  X⁻           ΔE = Eea

ΔE is change in energy

Eea is electron affinity

Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:

1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.

2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.

3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.

Applying these consideration to the elements given in the question:

(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)

(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)

(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)

(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)

(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)

The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².

An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Typically, these atoms tend to have half-filled or filled subshell configurations. In the given examples, the atom with electron configuration 1s2 2s2 2p6 3s2 3p1 is likely to have a positive electron affinity.

The electron affinity of an atom is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Usually, the atom would tend to have a half-filled (e.g. 3p3) or filled (e.g. 3p6) subshell configuration.

If we consider the given electron configurations: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5, it appears that (iii) 1s2 2s2 2p6 3s2 3p1 should have a positive electron affinity. This atom is a Potassium atom (K). It has a positive electron affinity because it is just one electron away from obtaining a full orbit shell (in the 3s and 3p orbitals) which would provide a more stable electron configuration.

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Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

[tex]a_{t}=r\alpha[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}[/tex]

[tex]a_{t}=0.0301\ m/s^2[/tex]

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

We know the equation

            [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

   where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

              [tex]g= \frac{4\pi^2l}{T^2}[/tex]

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

                     [tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

The initial velocity of the car is 36.6 m/s

Explanation:

The motion of the car is a uniformly accelerated motion (=constant acceleration), therefore we can apply suvat equations:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem, we have:

v = 0 is the final velocity (the car comes to a stop)

[tex]a=-3.2 m/s^2[/tex] is the acceleration

s = 210 m is the displacement of the car

Solving for u, we find the initial velocity:

[tex]u=\sqrt{v^2-2as}=\sqrt{0-(2)(-3.2)(210)}=36.6 m/s[/tex]

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Answer:

Explanation:

The particles are in x-y plane with coordinates of masses as follows

m₂ at (0,0 ) m₁ at ( 0,2 ), m₄ at ( 2,2 ) and m₃ at (2,0 )

Moment of inertia about z axis

I_z = 0 + 3 x 2² + 4 x (2√2)² + 3 x 2²

= 12 + 32 + 12

= 56 kgm²

Now let us find out moment of inertia about axis through CM

According to theorem of parallel axis

I_z = I_g + m x r²  

Here m is total mass that is 14 kg and r is distance between two axis which is √2 m

56 = I_g + 14 x (√2)²

I_g  = 56 - 28

= 28 kgm²

We can directly compute  I_g  as follows

I_g = 4 x (√2)² +3 x (√2)² +4 x (√2)²+3 x (√2)²

= 8 +6 +8 +6

= 28 kgm²

So the result obtained earlier is correct.

Answer:

t = 8.5 s

Explanation:

Kinematic equation of the movement of the tennis ball that is thrown upwards :

y = y₀ + v₀*t -½ g*t²   Equation (1)

Where :  

y : position of the ball as a function of time

y₀ : Initial position of the ball

t: time  

g: acceleration due to gravity in m/s²

Known data  

g = 32 ft/s²

y₀ = 680 ft

v₀ = 56 ft/s

Calculation of the time it takes for the ball to thit the ground

We replace data en the equation (1)

y = y₀ + v₀*t -½ g*t²  

0 = 680+(56)*t -½( 32) *t²

16*t²-(56)*t- 680 = 0  equation (2)

solving equation (2) quadratic:

t₁ = 8.5 s

t₁ = -5 s

Time cannot be negative so the time it takes for the ball to hit the ground  is t = 8.5 s

Answer:

The answer to your question is 11.2 m/s

Explanation:

Data

Initial speed (vo) = 6.0 m/s

Acceleration (a) = 3.0 m/s²

time = 15 s

Final speed = ?

Formula

                d = vot + [tex]\frac{1}{2} at^{2}[/tex]

                vf² = vo² + 2ad

Process

                d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]

                d = 90 + 337.5

                d = 427.5 m

                vf² = (6)² + 2(3)(15)

                vf² = 36 + 90

                vf² = 126

                vf = 11.2 m/s

Answer:

The enthalpy for dissolution is - 305.558 J/g

Solution:

Mass of the ionic compound, m = 10.00 g

Mass of water, m' = 75.0 g

Initial temperature, T = [tex]23.2^{\circ}C[/tex]

Final Temperature, T' = [tex]31.8^{\circ}C[/tex]

Now,

To calculate the change in enthalpy:

We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]

Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g

Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]

Thus

The heat absorbed by the solution is given by:

[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]

Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]

Answer:

Qtot = 6C * deltaV

Explanation:

you can find the total capacitance from adding 1C+2C+3C=6C. and the total voltage is 1V. Capacitance = charge/voltage--> C = Q / V--> 6C = Q / deltaV. this makes Qtot = 6C* deltaV

The Total charge for the equivalent circuit is =  [tex]Q_{tot}[/tex] = 6c * ΔV

  Although your question is incomplete I found the missing part online and used it to resolve the question

Given data :

Total capacitance ( C ) = 6C ( 1 + 2 + 3 )

voltage = 1 V

Three capacitors having values of ; 1 C, 2 C,  3 C

Determine the total charge ( Qtot )

Applying the formula ; Q = CV ----  ( 1 )

 where; Q =  charge

              C = capacitance

              change in V = ΔV

∴ [tex]Q_{tot}[/tex] = 6c * ΔV

Hence the total charge Qtot for the equivalent capacitor =  6c * ΔV

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Answer:

[tex]P_c=P_b>P_a[/tex]

Explanation:

E = Energy

T = Time

Power is given by the equation

[tex]P=\frac{E}{T}[/tex]

For first case

[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]

For second case

[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]

For third case

[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]

The rank of power would be [tex]P_c=P_b>P_a[/tex]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × [tex]\frac{ \lambda }{2}[/tex]      ........1

so  λ = [tex]\frac{2L}{10}[/tex]  

and velocity is express as

V = [tex]\sqrt{\frac{T}{\mu } }[/tex]    .................2

so

frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]

f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]

f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]      

and

f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]

so

beat frequency is = f2 - f1

put here value

beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex]  - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]

beat frequency = 13.87 Hz

Answer:

the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction

Explanation:

Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.

v = u + at

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

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

where,

u = initial velocity

v = final velocity

s = displacement

a = acceleration

t = time

Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.

t = 1s

u = 0 (since the hammer is dropped)

v = -1.6m/s (since its direction is downward)

a = ?

The only equation that connects all the above quantities is

v = u + at

therefore,

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

substituting the values

a = [tex]\frac{-1.6 - 0}{1}[/tex]

a = -1.6m/[tex]s^{2}[/tex]

Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.

Answer:

Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]

Explanation:

We are given that  

Frequency of the electromagnetic radiation,  [tex]f=5.16\times 10^{16}[/tex] Hz

The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

[tex]K=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula  

Then, the work function of the metal is given by :

[tex]W_o=h\nu -K[/tex]

[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]

[tex]W_o=3.38\times 10^{-17}\ J[/tex]

So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.

Answer:

a)Time taken will be 2.783 s

b)Time taken will be 2.564 s

Explanation:

a)Since the pulley has mass ,

[tex]m_{1}g-T_{1}= m_{1}a//T_{2}-m_{2}g=m_{2}a[/tex] ------3

             αR = a ;                    ------1

[tex](T_{1} - T_{2})*R=m*R^{2}*(alpha)=5*R^{2}*\frac{a}{R}[/tex] ------2

solving these we get ,

[tex]a=\frac{4g}{33}[/tex]

∴

a)time taken :

[tex]s=ut+\frac{1}{2} at^{2}\\u=0\\a=\frac{4g}{33} \\s=4.6\\4.6=\frac{1}{2} * \frac{4*9.8}{33} *t^{2}\\t= 2.783 sec[/tex]

ANS : 2.783 sec

b)

In case of B, mass is zero.

Thus, there is no rotation of the pulley. this is equivalent to a normal 1 Dimension motion question

[tex]m_{1}g-T_{1}=m_{1}a\\T_{1}-m_{2}g=m_{2}a\\a= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g\\a=\frac{4g}{28} \\a=\frac{g}{7} \\a=1.4ms^{-2}[/tex]

Thus time t will be ,

[tex]s=\frac{1}{2} at^{2}\\4.6=\frac{1}{2}*1.4*t^{2}\\ t=2.564 sec[/tex]

ANS : 2.564 sec

The time it takes the block to reach the ground can be found by making

use of the law of conservation of energy.

Reasons:

Given parameters are;

Mass of block m₁ = 16.0 kg

Mass of block m₂ = 12.0 kg

Mass of the pulley, M = 5.00 kg

By conservation of energy, we have;

m₁g·h - m₂·g·h = 0.5×(m₁ + m₂)·v² + 0.5·I·ω²

[tex]\omega = \dfrac{v}{R}[/tex]

[tex]Moment \ of \ inertia\ of \ pulley, I = \dfrac{1}{2} \cdot M \cdot R^2[/tex]

Therefore;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2 + 0.5 \cdot I \cdot \dfrac{v}{R}[/tex]

Which gives;

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v² + 0.5×(0.5×5×0.3²)× [tex]\left(\dfrac{v}{0.3} \right)^2[/tex]

Solving gives, v ≈ 21.93 m/s

We have;

v ≈ 3.544

v² = 2·a·h

[tex]a = \dfrac{v^2}{2 \times h}[/tex]

Which gives;

[tex]a = \dfrac{3.544^2}{2 \times 4.6} \approx 1.365[/tex]

v = a×t

[tex]t = \dfrac{v}{a} = \dfrac{3.544}{1.365} \approx 2.596[/tex]

The time it takes the the block m₁ to hit the floor, t ≈ 2.596 seconds

b. When the mass of the pulley is neglected, we have;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2[/tex]

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v²

180.504 = 14·v²

[tex]v = \sqrt{\dfrac{180.504}{14} } \approx 3.591[/tex]

[tex]a = \dfrac{3.591^2}{2 \times 4.6} \approx 1.401[/tex]

[tex]t = \dfrac{v}{a} = \dfrac{3.591}{1.401} \approx 2.562[/tex]

The time it takes the the block m₁ to hit the floor,  if the mass of the pulley were neglected, t ≈ 2.562 seconds.

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Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}[/tex]

[tex]\frac{0.877GMm}{R} =\frac{-GMm}{R+h}[/tex]

h = .14 R

b )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}[/tex]

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0