An object undergoing simple harmonic motion takes 0.34 s to travel from one point of zero velocity to the next such point. The distance between those points is 27 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.

Answer:

Yeah, it can be diffracted. Though it depends on a diffracting medium.

It must have some magnetic fields .

Forexample; X-ray diffraction where electrons are diffracted to the target filament.

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

Learn more about the ideal gas equation here:

https://brainly.com/question/18518493

Answer:

EMF =  73.5 volts

Explanation:

Given that,

The number of a coil, N = 42

The area of the coil, A = 0.125 m²

It is stretched to have no area in 0.100 s

The magnetic field strength is 1.4 T.

We need to find the average induced emf in the coil. We know that,

[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]

[tex]\epsilon=N\dfrac{BA}{dt}\\\\\epsilon=42\times \dfrac{1.4\times 0.125}{0.1}\\\\=73.5\ V[/tex]

So, the induced emf in the coil is 73.5 volts.

Answer:

Approximately [tex]63\; \rm m[/tex].

Explanation:

Convert the initial speed of this truck to standard units:

[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].

Calculate the initial kinetic energy of this truck:

[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].

The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.

Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:

[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].

This result is negative because the direction of friction is opposite to the direction of the motion of the truck.

Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].

Answer:

(a) Altitude = 1.95 x 10⁶ m = 1950 km

(b) g = 5.9 m/s²

Explanation:

(a)

The time period of the satellite is given by the following formula:

[tex]T^2 = \frac{4\pi^2r^3}{GM_E}[/tex]

where,

T = Time period = (125 min)([tex]\frac{60\ s}{1\ min}[/tex]) = 7500 s

r = distance of satellite from the center of earth = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

[tex]M_E[/tex] = Mass of Earth = 6 x 10²⁴ kg

Therefore,

[tex](7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\[/tex]

r = 8.29 x 10⁶ m

Hence, the altitude of the satellite will be:

[tex]Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m[/tex]

Altitude = 1.95 x 10⁶ m = 1950 km

(b)

The weight of the satellite will be equal to the gravitational force between satellite and Earth:

[tex]Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}[/tex]

g = 5.9 m/s²

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Answer:

Time to move out of the way = 1.74 s

Explanation:

Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.

Time to move out of the way = 1.74 s

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V

Answer:

(a) v = 32 t

h = 16 t^2

g = 32 ft/s^2

(b) 64 ft/s

Explanation:

height, h = 64 feet

g = - 32 ft/s^2

(a) Let the time is t .

Let the velocity after time t is v.

Use first equation of motion

v = u + at

- v = 0 - 32 t

v = 32 t

Let the distance is h from the top.

Use second equation of motion

[tex]h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2[/tex]

The acceleration is constant for entire motion.

(b) Let the velocity is v as it hits the ground. Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s[/tex]

Answer:

The fraction of kinetic energy to the total energy is [tex]\frac{K}{T}=\frac{8}{9}[/tex].

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

[tex]K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)[/tex]

The total energy is

[tex]T =0.5 mw^2A^2 ..... (2)[/tex]

Divide (1) by (2)

[tex]\frac{K}{T}=\frac{8}{9}[/tex]

Answer:

1.671L

Explanation:

In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.

Work done= external pressure× change in volume.

288= 2×( final volume-intial volume)×101.32

144÷101.32=(finalvolume-0.250)

1.421=final volume-0.250

final volume=1.671L

Answer:

The change in the internal energy of the first system is 300 J

The second system will do zero work in order to have the same internal energy.

Explanation:

Given;

heat added to the first system, Q₁ = 500 J

heat added to the second system, Q₂ = 300 J

work done by the first system, W₁ = 200 J

The change in the internal energy of the system is given by the first law of thermodynamics;

ΔU = Q - W

where;

ΔU is the change in internal energy of the system

The change in the internal energy of the first system is calculated as;

ΔU₁ = Q₁ - W₁

ΔU₁ = 500 J - 200 J

ΔU₁ = = 300 J

The work done by the second system to have the same internal energy with the first.

ΔU₁ = Q₂ - W₂

W₂ = Q₂ - ΔU₁

W₂ = 300 J - 300 J

W₂ = 0

The second system will do zero work in order to have the same internal energy.

Answer:

 y = 1.19 m  and λ = 8.6036 10⁻⁷ m

Explanation:

This is a slit interference problem, the expression for destructive interference is

          d sin θ = m λ

indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m

          λ = d sin  θ / m

let's calculate

          λ = 4.50 10⁻⁶ sin 35  /3

          λ = 8.6036 10⁻⁷ m

for the separation distance from the central stripe, we use trigonometry

         tan θ= y / L

         y = L tan θ

the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits

         L = 2 -0.30

         L = 1.70 m

let's calculate

        y = 1.70 tan 35

        y = 1.19 m

• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C

• Potential (V) = 1000 V

• Distance (r) = ?

[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]

[tex]\implies r= \dfrac{KQ}{V}[/tex]

[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]

[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]

[tex]\implies\bf r= 90\:m \\ [/tex]

Hence,the option B) 90 m is the correct answer.

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

[tex]t_{min[/tex] = ( λ / 4n )

so we simply substitute in our given values;

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m

[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m

[tex]t_{min[/tex] = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Answer:

[tex]r=200m[/tex]

Explanation:

From the question we are told that:

Charges:

[tex]Q_1=8.0mC[/tex]

[tex]Q_2=2.0mC[/tex]

[tex]Q_3=8.mC[/tex]

Distance [tex]d=300m[/tex]

Generally the equation for Force is mathematically given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Therefore

[tex]F_{32}=F_{31}[/tex]

[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]

[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]

[tex]r=2(300-r)[/tex]

[tex]r=200m[/tex]

Answer:

[tex]\omega_1=150rads/sec[/tex]

Explanation:

From the question we are told that:

Number of Revolution [tex]N=15=30\pi[/tex]

Deceleration [tex]d= -120 rads/2[/tex]

Generally the equation for  initial angular speed [tex]\omega_1[/tex] is mathematically given by

 [tex]\omega_2^2=\omega_1^2 +2(d)(N)[/tex]

 [tex]0=\omega_1^2 +2(-120)(20 \pi)[/tex]

 [tex]\omega_1^2=7200 \pi[/tex]

 [tex]\omega_1=150rads/sec[/tex]

Answer:

12.74 V

Explanation:

We are given that

Current, I1=54 A

Potential difference, V1=9.18V

I2=2.10 A

V2=12.6 V

We have to find the battery's emf.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=9.18+54r[/tex] ....(1)

[tex]E=12.6+2.10r[/tex]  .....(2)

Subtract equation (1) from (2)

[tex]0=3.42-51.9r[/tex]

[tex]3.42=51.9r[/tex]

[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]

Using the value of r in equation (1)

[tex]E=9.18+54(0.0659)[/tex]

[tex]E=12.74 V[/tex]

Answer:

The correct option is (B).

Explanation:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric potential due to a point charge is given by :

[tex]V=\dfrac{kQ}{r}[/tex]

Where

k is the electrostatic constant

Q is the electric charge

r is the distance from the charge

So, a positive electric charge in a region of changing electric potential will move in the direction of increasing potential.

Answer:

C is the right answer

Explanation:

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Answer:

[tex]M=7.05*10^{-4}[/tex]

Explanation:

From the question we are told that:

Coil one turns N_1=1550 Turns/m

Radius [tex]r=0.0240m[/tex]

Turns 2 [tex]N_2=200N[/tex]

Generally the equation for area is mathematically given by

[tex]A=\pi*r^2[/tex]

[tex]A=\pi*0.024^2[/tex]

[tex]A=\1.81*10^{-3} m^2[/tex]

Therefore

The mutual inductance of this system is

[tex]M=\mu*N_1*N_2*A[/tex]

[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]

[tex]M=7.05*10^{-4}[/tex]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]

Therefore, the maximum speed is 0.20 m/s

Answer:

The correct answer is "2205.72 N".

Explanation:

Given:

m₁ = 161.5 kg

m₂ = 72.3 kg

By taking moment about point A, we get

⇒ [tex]m_1 g Cos \theta.\frac{l}{2 Cos \theta} + m g Cos \theta.\frac{l}{ Cos \theta} -T Cos(90-2 \theta).\frac{l}{2 Cos \theta} = 0[/tex]

By substituting the values, we get

⇒ [tex]\frac{161.5\times 9.8\times 13}{2}+72.3\times 9.8\times 13-T.2 Sin \theta.l =0[/tex]

⇒ [tex]10287.55+9211.02-T 2\times 0.34\times 13=0[/tex]

⇒                                            [tex]T\times 8.84=19498.57[/tex]

⇒                                                       [tex]T= 2205.72 N[/tex]

Answer:

Explanation:

Let the mass of objects be m₁ and m₂ .

Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²

Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4  =  1/2 ( m₁ + m₂ ) v² x .25

final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²

= 0.25

b )

Applying law of conservation of momentum to the system . Let m₁ > m₂

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁  -  m₂  = ( m₁ + m₂ )  / 2

2m₁ - 2 m₂ = m₁ + m₂

m₁ = 3m₂

m₁ / m₂ = 3 / 1

Answer:

(a) The ratio is 1 : 4.

(b) The ratio is 1 : 3.

Explanation:

Let the mass of each object is m and m'.

They initially move with velocity v opposite to each other.

Use conservation of momentum

m v - m' v = (m + m') v/2

2 (m - m')  = (m + m')

2 m - 2 m' = m + m'

m = 3 m' .... (1)

(a) Let the initial kinetic energy is K and the final kinetic energy is K'.

[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]

[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]

The ratio is

K' : K = 1 : 4

(b) m = 3 m'

So, m : m' = 3 : 1

Answer:

e. TA>T>Tc

Explanation:

a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:

[tex]Q=mC_{p}(T_{f}-T_{0})[/tex]

if we did an energy balance of the system, we would get that>

QA+QB+QC=0

For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.

We don't know either of them, so we cannot determine if this statement is true.

b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.

c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.

d) is not complete

e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.

Answer:

   1500 W to 2200 W

Explanation:

Every person does work in his daily day to day life. A person needs energy in order to perform work. And the energy consumed by an individual while performing a daily work is directly responsible to the mount of oxygen consumed by the person.

The USDA is the federal agency which looks after the food and agriculture matters of the US government. It deals with and formulates different policies and laws for the country and it s people. It recommends about 2000 calories per day for women and for men it recommends about 2500 calorie per days of food intake.

Accordingly, the average power required by a human body for doing regular work is in the range of 1500 W to 2200 W.

Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.

Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.

Answer:

Explanation:

Moment of inertia of sphere I = 2 /5 x m R²

m is mass and R is radius of the sphere .

I = 2/5 x 1800 x 0.5²

= 180 kg m².

For rotational motion starting from rest

θ = 1/2 α t²

Here θ = π /2 = 1.57 radians

t = 12 s

1.57 radians  = 1/2 x α x 12²

Angular acceleration α = .022 rad /s²

Torque applied = I α

=  180 kg m² x  .022 rad /s²

= 3.96 N.m

b ) Torque = F x r

F = Torque / r

= 3.96 / .5 = 7.92 N .

c )

Average weight of a cellophone is 1.3  N .

So the force applied is a little more  than weight of a cellophone.