A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing

Answer:

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

[tex]g[/tex] - Gravitational constant, in meters per square second.

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Answer:

Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.

Divide by 100.

Multiply by proper voltage drop value in tables. The result is voltage drop.

Explanation:

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

Answer:

Correct option not indicated

Explanation:

There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).

The formula to calculate a centripetal force (F) is

F = mv²/r

Where m is mass, v is velocity and r is radius

where

While the formula to calculate a centrifugal force (F) is

F = mω²r

where m is mass, ω is angular velocity and r is radius of the circular path.

From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.

NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

For more information on Charge visit

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Answer:

The frequency of the second wave is half of the frequency of first one.

Explanation:

The wavelength of the second wave is double is the first wave.

As we know that the frequency is inversely proportional to the wavelength of the velocity is same.

velocity = frequency x wavelength

So, the ratio of frequency of second wave to the first wave is

[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]

The frequency of the second wave is half of the frequency of first one.

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]

Electric field, [tex]E=55\ N/C[/tex]

We need to find the distance. We know that, the electric field a distance r is as follows :

[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]

So, the required distance is 20.22 m.

Answer:

The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).

Explanation:

We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.

P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.

Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)

Answer:

F = 1.638 x 10⁸ N = 163.8 MN

Explanation:

The total force exerted by the molasses is given as:

F = PA

where,

F = Force exerted by the molasses = ?

P = Pressure = ρgh

ρ = density of molasses = 1600 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height of tank = 17.7 m

A = cross-sectional area of tank = πr²

r = radius of tank = 27.4 m/2 = 13.7 m

Therefore,

[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]

F = 1.638 x 10⁸ N = 163.8 MN

Answer:

fair palybtgshsisuehdh

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Explanation:

a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy

b) burning of coal converts chemical energy to heat energy

c) rubbing of both hands against each other converts mechanical to heat energy

Answer:

a. solar cells

b.coal,wood,petroleum

c.rubbing ours palms

Answer:

Explanation:

Apply law of conservation of momentum along y-axis.

Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.

Let the mass of each piece after breaking be m .

Momentum of piece moving along positive y-axis

= m x 10 = 10m .

Let the component of velocity of third piece along y-axis be v .

Its momentum along the same direction = m v .

Total momentum along y -axis = 10 m + m v

According to law of conservation of momentum

10 m + mv = 0

v = - 10 m/s .

Component of velocity of the third piece along y-axis will be - 10 m/s .

In other words it will be along negative y-axis with speed of 10 m/s.

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex]I = \frac{ E}{ R + r} \\[/tex]

[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]

Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have

[tex]➺ \: I \: (R + r) = E \times 1[/tex]

[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]

[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]

[tex]\boxed{ Hence\:proved. }[/tex]

[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities