A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. From the center of one particular fringe to the center of the ninth bright fringe from this one is 1.6 cm. What is the wavelength of the light? [530nm]

Answer:

600N

Explanation:

If the ball bounces off upward with the same speed, the velocities before and after are equal and in opposite directions. That means the floor must have caused a momentum impact to change its direction.

By the law of momentum conservation:

[tex]mv + \Delta P_f = mV[/tex]

where m = 3 kg is the ball mass. v = -12 m/s is the velocity right before the impact. V = 12m/s is the velocity right after the impact. ΔP_f is the momentum caused by the floor.

[tex]\Delta P_f = m(V-v) = 3(12 - (-12)) = 3*24 = 72kgm/s[/tex]

Then the average force exerted during the 0.12s impact is

[tex]F = \frac{\Delta P_f}{\Delta t} = \frac{72}{0.12} = 600N[/tex]

Answer: 0.47 rad/sec

Explanation:

By definition, the angular velocity is the rate of change of the angle traveled with time, so we can state the following:

ω = ∆θ/ ∆t

Now, we are told that in 13.3 sec, the ball completes one revolution around the circle, which means that, by definition of angle, it has rotated 2 π rad (an arc of 2πr over the radius r), so we can find ω as follows:

ω = 2 π / 13.3 rad/sec = 0.47 rad/sec

Answer:

0.472rad/s

Explanation:

Angular velocity = 2πf where f = frequency and frequency is the number of revolution per second and 2π represent a cycle of revolution. The mass of the body was 5kg, the time taken to complete a cycle was 13.3 s.

Frequency = 1/period where period is the time it takes to complete a revolution.

F = 1/13.3 = 0.075hz

Angular velocity = 2* 3.142* 0.075 = 0.472rad/s

If a person pulls ten feet of the rope tied to a vehicle through a pulley, the vehicle will move the same distance, which is ten feet assuming ideal conditions with no slack or stretch in the system.

The question is asking about how far a vehicle moves when a rope attached to it and run through a pulley is pulled a certain distance. This scenario can be understood using the principles of distance and leverage as applied in physics. When you pull a rope through a pulley, the movement or distance covered by the object attached to the rope is equivalent to the length of the rope pulled.

In this case, if a person pulls ten feet of the rope, the vehicle will also move ten feet. It's important to note that this is applicable under ideal conditions where the rope and pulley do not stretch or bend and the there is no slack in the system.

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Answer:

3.6 kHz

Explanation:

The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.

The first harmonic will be as seen in the figure attached.

The length of the first harmonic will be λ/4.

λ/4=2.4 cm

λ=2.4 * 4=9.6 cm 0.096 m

Speed of Sound- 344 m/s(in air)

velocity(v) * Time Period(T) = Wavelength (λ)

Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}

\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency

Plugging in the values into the equation,

Frequency = [tex]\frac{344}{0.096}[/tex] Hz

                  = 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

Answer:

3.6 kHz

Explanation:

The auditory canal is a closed pipe because it has one closed end, the end terminated at the eardrum.

The length of the first harmonic of a closed pipe is given as;

L = λ/4 --------------------  (i)

where L = Length

and λ = wavelength

2.4 = λ/4

λ = 2.4 x 4 = 9.6 cm

Also, v = fλ  ------------------ (ii)

where v = speed of sound in air = 344 m/s

f = frequency of wave in Hertz

f = v/λ ------------------ (iii)

convert 9.6 cm to m = 0.096 m

substitute for λ and v in (iii)

[tex]f = \frac{344}{0.096} = 3583.33[/tex]

3583.33 Hz = 3600 Hz = 3.6 kHz

Answer: 3. The puck B remains at the point of collision.

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.

The initial momentum is due only to puck B, as Puck A is at rest.

The final momentum is given by the sum of the momenta of both pucks, so we can write the following equation:

mA*viA = (mA * vfA) + (mB * vfB)

As mA = mB = m, we can simplify the former equation as follows:

viA = vfA + vfB   (1)

Now, we also know, that the collision was an elastic collision, so total kinetic energy must be conserved too:

½ m viA²= ½ m vfA² + ½ m vfB²

Simplifying on both sides, we finally have:

viA² = vfA² + vfB²   (2)

Now, if we square both sides of (1), we get:

viA² = (vfA + vfB)²= vfA² + 2* vfA * vfB  +vfB² (3)

As the right side in (2) and (3) must be equal each other (as the left sides do), the only choice is that either vfA or vfB, be zero.

As we are told that puck A (initially stationary) after the collision, moves, the only possible choice is that puck B remain at rest in the point of collision, after the collision, exchanging his speed with puck A.

In an elastic collision where the total energy of the system is conserved, Puck B rebounds. This is because kinetic energy, along with momentum, is conserved in such a collision, causing the moving puck to rebound.

Considering this scenario from the lens of physics, the principle of conservation of momentum and kinetic energy comes into play, which means the total momentum and kinetic energy before and after the collision should remain constant. This is consistent with the properties of an elastic collision. Both pucks have the same mass and initially, Puck B is in motion while Puck A is stationary. Hence, all of the initial energy and momentum are with Puck B.

Upon collision, the energy is transferred, causing Puck A to move in the same direction that Puck B was initially moving. The correct statement is that if the collision is elastic and the total energy of the system is conserved, 'Puck B rebounds'. This rebounding comes from the conservation of kinetic energy principle which holds in an elastic collision. That is, the total initial kinetic energy equals the total final kinetic energy.

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Answer:

he did not work on the create

Explanation:

Honestly I have no idea it makes a lot of sense to use the work equation so that was my first answer but for some reason on UT Quest it said that was the wrong and moe not doing any work on the create was the right answer

Final answer:

The work done by Moe on the crate is 10,000 J.

Explanation:

The work done by Moe on the crate can be calculated using the formula:

Work = Force x Distance x Cosine(theta)

Since Moe is exerting a force of 1000 N and the distance covered is 10 meters, we can plug these values into the formula:

Work = 1000 N x 10 m x Cosine(0°)

The angle theta is 0° because Moe is exerting a force in the same direction as the displacement.

Cosine(0°) is equal to 1, so the work done by Moe on the crate is:

Work = 1000 N x 10 m x 1 = 10,000 J

Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}[/tex]

f = frequency (60 Hz)

c= capacitor (10 µF = [tex]10^-6[/tex])  

[tex]\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }[/tex]

Substitute the given values

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}[/tex]

[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}[/tex]

[tex]\mathrm{x}_{\mathrm{c}}=265.39 \Omega[/tex]

Given that, R = 200 Ω

[tex]X^{2}=R^{2}+X c^{2}[/tex]

[tex]X^{2}=200^{2}+265.39^{2}[/tex]

[tex]X^{2}=40000+70431.85[/tex]

[tex]X^{2}=110431.825[/tex]

[tex]x=\sqrt{110431.825}[/tex]

X = 332.31 Ω

[tex]\text { Current }(I)=\frac{V}{R}[/tex]

[tex]\text { Current }(I)=\frac{120}{332.31}[/tex]

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,  

[tex]\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200[/tex]

[tex]\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200[/tex]

[tex]\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}[/tex]

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12  

Answer:F=40.09 N

Explanation:

Given

weight of crate [tex]W=50 N[/tex]

Inclination [tex]\theta =37^{\circ}[/tex]

Frictional Force [tex]f=10 N[/tex]

as the crate is moving with constant velocity therefore net Force on crate is zero

[tex]F-50\sin (37)-f=0[/tex]

[tex]F=50\sin (37)+10[/tex]

[tex]F=30.09+10[/tex]

[tex]F=40.09 N[/tex]

Answer:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

    [tex]\lambda_{max}[/tex] T = 2,898 10⁻³ [m K]

    T = 2,898 10⁻³ / [tex]\lambda_{max}[/tex]

a) indicate that the wavelength is

    Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m

    Lam = 6.50 10⁻⁷ m

    T = 2,898 10⁻³ / 6.50 10⁻⁷

    T = 4,458 10³ K

    T = 4458K

b) lam = 570 nm = 5.70 10⁻⁷ m

    T = 2,898 10⁻³ / 5.70 10⁻⁷

    T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

    T = 2,898 10⁻³ / 4.7 10⁻⁷

    T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

    T = 2,898 10⁻³ / 5.20 10⁻⁷

    T = 5573K

A ) the surface temperature of red star is about 4500 K

B ) the surface temperature of yellow star is about 5100 K

C ) the surface temperature of blue star is about 6200 K

D ) the surface temperature of white star is about 5600 K

[tex]\texttt{ }[/tex]

Let's recall the Wien's Displacement Law as follows:

[tex]\boxed {\lambda_{max}\ T = 2.898 \times 10^{-3} \texttt{ m.K}}[/tex]

where:

λ_max = the wavelength of the maximum radiation energy ( m )

T = surface temperature of the star ( K )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

wavelength of red light = λ_r = 650 nm = 650 × 10⁻⁹ m

wavelength of yellow light =  λ_y = 570 nm = 570 × 10⁻⁹ m

wavelength of blue light =  λ_b = 470 nm = 470 × 10⁻⁹ m

wavelength of white light =  λ_w = 520 nm = 520 × 10⁻⁹ m

Asked:

A ) the surface temperature of red star = T_r = ?

B ) the surface temperature of yellow star = T_y = ?

C ) the surface temperature of blue star = T_b = ?

D ) the surface temperature of white star = T_w = ?

Solution:

[tex]T_r = ( 2.898 \times 10^{-3} ) \div \lambda_r[/tex]

[tex]T_r = ( 2.898 \times 10^{-3} ) \div ( 650 \times 10^{-9} )[/tex]

[tex]\boxed {T_r \approx 4500 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_y = ( 2.898 \times 10^{-3} ) \div \lambda_y[/tex]

[tex]T_y = ( 2.898 \times 10^{-3} ) \div ( 570 \times 10^{-9} )[/tex]

[tex]\boxed {T_y \approx 5100 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_b = ( 2.898 \times 10^{-3} ) \div \lambda_b[/tex]

[tex]T_b = ( 2.898 \times 10^{-3} ) \div ( 470 \times 10^{-9} )[/tex]

[tex]\boxed {T_b \approx 6200 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]T_w = ( 2.898 \times 10^{-3} ) \div \lambda_w[/tex]

[tex]T_w = ( 2.898 \times 10^{-3} ) \div ( 520 \times 10^{-9} )[/tex]

[tex]\boxed {T_w \approx 5600 \texttt{ K} }[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Mathematics

Chapter: Energy

Answer:

45500 J

Explanation:

Pressure, P =350 kPa = 350 x 1000 Pa

V1 = 0.02 m^3

V2 = 0.15 m^3

Work done by the piston

W = Pressure x increase in volume

W = P x (V2 - V1)

W = 350 x 1000 (0.15 - 0.02)

W = 45500 J

Thus, the work done is 45500 J.

Final answer:

The piston performs 45500 Joules of work on the gas when the volume of the gas in the piston-cylinder apparatus increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa.

Explanation:

To determine the work done by the piston on the gas when the cylinder volume increases from 0.02 m³ to 0.15 m³ at a constant pressure of 350 kPa, we can use the formula for work done during a quasi-static process in thermodynamics, which is W = PΔV, where W is the work done by the gas, P is the constant pressure, and ΔV is the change in volume.

Given:
P = 350 kPa = 350000 Pa (since 1 kPa = 1000 Pa)
ΔV = 0.15 m³ - 0.02 m³ = 0.13 m³

We can now calculate the work done:

W = PΔV
W = 350000 Pa × 0.13 m³
W = 45500 J

Therefore, the piston performs 45500 Joules of work on the gas when the volume increases to 0.15 m³.

Answer:(a)12.17 rad/s

Explanation:

Given

Moment of Inertia [tex]I=0.23 kg.m^2[/tex]

Torque [tex]T=2.8 N-m[/tex]

(a)Torque is given by Product of Moment of inertia and angular acceleration

[tex]T=I\cdot \alpha [/tex]

[tex]2.8=0.23\cdot \alpha [/tex]

[tex]\alpha =\frac{2.8}{0.23}=12.17 rad/s[/tex]

(b)RPM of blades [tex]N=205 rpm [/tex]

angular velocity [tex]\omega =\frac{2\pi N}{60}[/tex]

[tex]\omega =\frac{2\pi 205}{60}=21.47 rad/s[/tex]

Rotational Kinetic Energy [tex]=\frac{I\omega ^2}{2}[/tex]

[tex]=\frac{0.23\times (21.47)^2}{2}=53.01 J[/tex]

The angular acceleration of the blades is approximately 12.1739 rad/s^2. When the blades rotate at 205 rpm, the rotational kinetic energy is approximately 0.0948 J.

(a) To find the angular acceleration of the blades, we can use the formula:

torque = moment of inertia × angular acceleration

Plugging in the given values:

torque = 2.8 N · m

moment of inertia = 0.23 kg · m2

Rearranging the formula, we get:

angular acceleration = torque / moment of inertia

Substituting the values:

angular acceleration = 2.8 N · m / 0.23 kg · m2

Solving for angular acceleration gives us:

angular acceleration ≈ 12.1739 rad/s2

(b) To find the rotational kinetic energy, we can use the formula:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Plugging in the given values:

moment of inertia = 0.23 kg · m2

angular velocity = 205 rpm = 205 revolutions / 60 seconds = 3.4167 rev/s

Rearranging the formula, we get:

rotational kinetic energy = ½ × moment of inertia × (angular velocity)2

Substituting the values:

rotational kinetic energy = ½ × 0.23 kg · m2 × (3.4167 rev/s)2

Solving for rotational kinetic energy gives us:

rotational kinetic energy ≈ 0.0948 J

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To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

[tex]n_1sin\theta_1 = n_2sin\theta_2[/tex]

Where,

[tex]\theta =[/tex] Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

[tex]n_1 sin\theta_1 = n_2sin\theta_2[/tex]

[tex](1.54) sin\theta_1 = (1.33)sin(90)[/tex]

[tex]sin\theta_1 = \frac{1.33}{1.54}[/tex]

[tex]\theta = sin^{-1}(\frac{1.33}{1.54})[/tex]

[tex]\theta = 59.72\°[/tex]

Therefore the [tex]\alpha_{max}[/tex] would be equal to

[tex]\alpha = 90\°-\theta[/tex]

[tex]\alpha = 90-59.72[/tex]

[tex]\alpha = 30.27\°[/tex]

Therefore the largest value of the angle α is 30.27°

The largest value the angle α can have without any light being refracted out of the prism at face AC is 30.27°.

From the information given, the total internal reflection would be:

1.54sinb = (1.33) sin90°

sin b = (1.33 sin90° / 1.54)

b = 59.72°

Therefore, the value of the angle will be:

= 90° - 59.72°

= 30.27°

In conclusion, the correct option is 30.27°.

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Answer:

a) C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]

b) [tex](x_4,y_4)=(-1.917,-1.75)m[/tex]

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

[tex]\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}[/tex]

[tex]\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}[/tex]

Where M represent the sum of all the masses on the system.

And the center of mass C.M [tex]=(\bar x, \bar y)[/tex]

Part a

[tex]m_1= 3 kg, m_2=5kg,m_3=7kg[/tex] represent the masses.

[tex](x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)[/tex] represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m[/tex]

[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m[/tex]

C.M [tex]=(\bar x, \bar y)=(0.767,0.7)m[/tex]

Part b

For this case we have an additional mass [tex]m_4=6kg[/tex] and we know that the resulting new center of mass it at the origin C.M [tex]=(\bar x, \bar y)=(0,0)m[/tex] and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

[tex]\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m[/tex]

If we solve for a we got:

[tex](3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0[/tex]

[tex]a=-\frac{(5kg*2.3m)}{6kg}=-1.917m[/tex]

[tex]\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m[/tex]

[tex](3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0[/tex]

And solving for b we got:

[tex]b=-\frac{(7kg*1.5m)}{6kg}=-1.75m[/tex]

So the coordinates for this new particle are:

[tex](x_4,y_4)=(-1.917,-1.75)m[/tex]

Answer:

[tex]8.2\times 10^{15}\ m[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength = 600 nm

d = Diameter of mirror = 42 m

D = Distance of object

x = Diameter of Jupiter = [tex]1.43\times 10^8\ m[/tex]

Angular resoulution is given by

[tex]\Delta\theta=1.22\frac{\lambda}{d}\\\Rightarrow \Delta\theta=1.22\frac{600\times 10^{-9}}{42}\\\Rightarrow \Delta\theta=1.74286\times 10^{-8}\ rad[/tex]

We also have the relation

[tex]\Delta\theta\approx=\frac{x}{D}\\\Rightarrow D\approx\frac{x}{\Delta\theta}\\\Rightarrow D\approx\frac{1.43\times 10^8}{1.74286\times 10^{-8}}\\\Rightarrow D\approx 8.2049\times 10^{15}\ m[/tex]

The most distant Jupiter-sized planet the telescope could resolve is [tex]8.2\times 10^{15}\ m[/tex]

Answer:

a) I = 13.77 A

b) 0 ° or to the East

Explanation:

Part a

The magnetic field by properties would be 0 at the radius on this case r =8.1 cm.Analyzing the situation the wirde would produce a magnetic field equals in magnitude to the magnetic field on Earth by with the inverse direction.

The formula for the magnetic field due to a wire with current is:

[tex] B = \frac{\mu_0 I}{2 \pi r} [/tex]

In order to have a value of 0 for the magnetic field at the radius then we need to have this balance

B (r=8.1) = B (Earth)

Replacing:

[tex] B = \frac{\mu_0 I}{2 \pi r)}= B_{Earth} [/tex]

Solving from I, from the last equation we got:

[tex] I = \frac{2 \pi r B_{earth}}{\mu_0} [/tex]

[tex] I=\frac{2 \pi 0.081 m (34 x 10^{-6} T)}{4 \pi x 10^{-7} Tm/A}[/tex] = 13.77 A

Part b

We can use the right hand rule for this case.

The magnetic field of the wire would point to the South, because the magnetic field of the earth given points to the North. Based on this the current need's to flow from West to East in order to create a magnetic field pointing to the south, because the current would be perpendicular to the magnetic field created.

Answer:

The final kinetic energy is 242 J.

Explanation:

Hi there!

According to the work-energy theorem, the total work done on the box is equal to its change in kinetic energy:

W = ΔKE

Where:

W = work done on the box.

ΔKE = change in kinetic energy (final KE - initial KE).

The only forces that do work in this case are the applied force and the friction force because the box moves only horizontally.

The equation of work is the following:

W = F · s

Where:

F = force.

s = traveled distance.

Then, the work done by the applied force is:

W = 122 N · 9.46 m = 1.15 × 10³ J

To calculate the work done by friction, we have to find the friction force:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of kinetic friction.

The box does not have a net vertical acceleration. It means that the sum of the vertical forces acting on the box is zero:

∑Fy = 0

In this case, the only vertical forces are the weight of the box and the normal force. Then:

Weight + N = 0

N = - Weight

The weight of the box is calculated as follows:

Weight = m · g

Where:

m = mass of the box.

g = acceleration due to gravity.

Then:

-Weight = N = 28 kg · 9.8 m/s² = 274.4 N

Now, we can calculate the friction force:

Fr = N · μ

Fr = 274.4 N · 0.35 = 96 N

The work done by the friction force will be:

W = Fr · s

W = 96 N · 9.46 m = 908 J

Since the work done by friction opposes to the sense of movement, the work is negative.

Now, we can calculate the total work done on the box:

W total = W applied forece + W friction force

W = 1.15 × 10³ J - 908 J = 242 J

Applying the work-energy theorem:

W = final KE - initial KE

Since the box is initially at rest the initial kinetic energy is zero. Then:

W = final KE - 0

W = final KE

Final KE = 242 J

The final kinetic energy is 242 J.

Answer:

option A.

Explanation:

The correct answer is option A.

Two rocks are off a bridge first rock is fallen 4  when the second rock is dropped.

Both the rock is dropped under the effect of acceleration due to gravity so, the rate of change of the velocity for both the rock particle will be the same.

Hence, the first rock will reach ground earlier than the second rock because the rate of change of both the rock is at the same rate.

At the time when the two rocks should continue to fall so their velocities should be increased.

The correct answer is option A.

When two rocks are off a bridge so here first rock is fallen 4 at the time when the second rock should be fallen. Both the rock should be fallen under the impact of acceleration because of gravity due to which the rate of change of the velocity for both the rock particle should be similar.

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Answer:

The Q₂ is 0.318 μC

Explanation:

The charge flows is the same on both, then:

[tex]V_{1} =\frac{kQ_{1} }{r_{1} } \\V_{2} =\frac{kQ_{2} }{r_{2} } \\Q_{2} =Q-Q_{1} \\\frac{kQ_{1} }{r_{1} } =\frac{k*(Q-Q_{1}) }{r_{2} } \\Q_{1} =\frac{\frac{Q}{r_{2} } }{(1/r_{1})+(1/r_{2}) }[/tex]

But:

[tex]\frac{1}{r_{1} } +\frac{1}{r_{2} }=\frac{1}{0.38} +\frac{1}{0.28} =6.2[/tex]

Q = = 0.75 μC

Replacing:

[tex]Q_{1} =\frac{\frac{0.75}{0.28} }{6.2} =0.432\mu C[/tex]

The Q₂ is equal:

Q₂ = 0.75 - 0.432 = 0.318 μC

The second conducting sphere, once connected and then disconnected from the first, ends up with the same charge as the first sphere, 0.75 µC or 0.75 x 10^-6 C.

When two conducting spheres are connected by a thin conducting wire, the charge distributes itself evenly across both spheres, assuming both spheres are identical in size. In the case of spheres with different radii, the amount of charge on each sphere once they are separated again will still be proportional to the original amount. Since the two spheres in the question are described to be separated by a large distance relative to their size, the thin wire connecting them would create an equipotential surface, allowing the charge to redistribute.

The total charge Q is conserved in the system, thus the charge on the second sphere Q2 after disconnecting the wire would be the same as the initial total charge Q since it was not charged before. Therefore, the total charge on sphere two, Q2, will also be 0.75 μC or 0.75 x 10^{-6} C.

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Answer:

Choice A. The bouncy "superball" will exert approximately twice as much force than the blob of clay when tossed out at the same speed.

Explanation:

The momentum of an object is equal to the product of its mass and its velocity. The momentum of the system is conserved even if the collision is inelastic.

When the bouncy ball or the blob of clay hits the door, their momentum changes. Since momentum is conserved, the door would move in the opposite direction such that the total momentum stays at zero.

Let the mass of each object be [tex]m[/tex]. Suppose that both objects hit the door at a speed of [tex]v[/tex] in the same direction.

By Newton's Second Law, the net force on the door is proportion to the rate of change in its momentum. Assume that the two objects are in contact with the door for the same amount of time, the bouncy ball would exert about twice as much force on the door as the clay would. Hence it is "more effective" for closing the door.

The clay is more effective than the superball for shutting the door because upon impact the clay sticks to the door and transfers all of its momentum to it, while the superball bounces back and doesn't transfer all of its momentum.

The most effective object to throw at your door to shut it would be c. The blob of clay. This is because of the principle of conservation of momentum. When the clay hits the door, it sticks to it, thus transferring all of its momentum to the door. This momentum is what causes the door to move and potentially close. On the other hand, when the superball hits the door, it bounces back, meaning not all of its momentum is transferred to the door. Hence, the clay, by virtue of its ability to stick and transfer its full momentum, would be more effective.

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Answer:

a) The central atom of the amide ion is nitrogen

b) (NH2)-

c) There are two lone pairs around the nitrogen atom

d) The ideal angle between nitrogen hydrogen bonds is 109.5° C, typical of a tetrahedral electron density arrangement

e) I would expect the nitrogen-hydrogen bond angles in the amide ion to be less than 109.5 ° because unbonded pairs repel bonded pairs much more than other bonded pairs do. So the bonds in this ion would be pushed closer together than normal

Answer:[tex]1.7\times 10[/tex]

Explanation:

Given

Force [tex]F=7 N[/tex]

time interval [tex]t=2.5 s[/tex]

Impulse is given by [tex]=Force \times time\ interval\ for\ applied\ force [/tex]

[tex]Impulse=7\times 2.5=17.5[/tex]

For two significant Figure

[tex]Impulse=1.7\times 10[/tex]

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) [tex]\omega^{2} =\frac{g}{L}[/tex]

where, L be the length of the pendulum

[tex]L =\frac{g}{\omega ^{2}}[/tex]

[tex]L =\frac{9.8}{9.2 \times 9.2}[/tex]

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

Answer:

Explanation:

initial velocity, u = 28 m/s

Angle of projection, θ = 30°

The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.

Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s

At t = 2 sec, the horizontal component of velocity = 24.25 m/s

At t = 3 sec, the horizontal component of velocity = 24.25 m/s

In this exercise we have to use the knowledge about oblique launch to calculate the components of velocity for each case, so we have that:

For all times we will find a velocity equal to 24.25 m/s

organizing the information given in the statement we have that:

Knowing that the component can be written as:

[tex]u cos \theta = 28 * Cos 30 = 24.25 m/s[/tex]

So as the formula does not depend on time we have that for any value it will have a constant velocity.

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Answer:

r = 0.31 m

Explanation:

Given that,

Mass of the sculpture, m = 191 kg

The sculpture's moment of inertia with respect to the pivot is, [tex]I=17.2\ kg-m^2[/tex]

Frequency of oscillation, f = 0.925 Hz

Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}[/tex]

[tex]r=\dfrac{4\pi^2f^2I}{mg}[/tex]

[tex]r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}[/tex]

r = 0.31 m

So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.

Answer:

0.04s

Explanation:

Specific heat of water c = 4.186J/g.C

The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is

[tex]E = mc\Delta T = 0.5 * 4.186 * (72 - 30) = 87.9 J[/tex]

If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is

[tex] t = \frac{E}{P} = \frac{87.9}{2200} = 0.04 s[/tex]

Answer:

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

[tex]x(t) = Acos\omega t[/tex]

where

A = Amplitude

[tex]\omega[/tex] = angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

[tex]\omega = \sqrt{k}{m}[/tex]

[tex]\omega^{2} = \frac{k}{m}[/tex]

Thus

[tex]k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m[/tex]

(c) Position after one half period:

[tex]x(t) = 0.42cos\pi = - 0.42\ m[/tex]

(d) After one third of the period:

[tex]x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m[/tex]

(e) Time taken to get at x = - 0.10 m:

[tex]-0.10 = 0.42cos6t[/tex]

[tex]6t = co^{- 1} \frac{- 0.10}{0.42}[/tex]

t = 17.3 s

The amplitude of oscillation is 0.42 m. The force constant of the spring is approximately 0.03 N/m. The position of the mass after one half a period, one-third of a period, and at x = -0.10 m can be determined using the given equation x(t).

The amplitude of oscillation is the maximum displacement from the equilibrium position. In this case, the amplitude is given as 0.42 m.

The force constant (k) of the spring can be determined using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from equilibrium. In this case, the force constant is calculated by dividing the mass (0.56 kg) by the square of the angular frequency (6 rad/s) squared, which gives a force constant of approximately 0.03 N/m.

The position of the mass after it has been oscillating for one half a period can be found by substituting the value of time (T/2) into the equation x(t), which gives a position of approximately -0.42 m.

The position of the mass one-third of a period after it has been released can be determined by substituting the value of time (T/3) into the equation x(t), which gives a position of approximately 0.33 m.

The time it takes the mass to get to the position x = -0.10 m after it has been released can be found by rearranging the equation x(t) and solving for time. The time is approximately 0.14 seconds.

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Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

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

[tex]A=\pi\times(2.00\times10^{-2})^2[/tex]

[tex]A=0.00125\ m^2[/tex]

[tex]A=1.25\times10^{-3}\ m^2[/tex]

We need to calculate the deformation

Using formula of deformation

[tex]\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)[/tex]

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

[tex]\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)[/tex]

[tex]\Delta x=0.000704\ m[/tex]

[tex]\Delta x=7.04\times10^{-4}\ m[/tex]

[tex]\DElta x=0.70\ mm[/tex]

Hence, The deformation in the pole due to force is 0.70 mm.

Answer:

e. 160.0 mm

Explanation:

Precision is defined as the accuracy in measurement that the measuring instrument measure any given specimen. The instrument having the lowest least count will be more precise than the instrument with higher least count.

Also precision of the reading is given by the number of the significant figures in the given sample of reading.

Option (e) that is 160.0 has a total number of significant figure of 4, which is the largest total significant figures among the other options.

Also the measurement reads 160.0 mm which measure to the lowest possible measurement of the given pen. Thus this reading is more precise.

Hence the correct option is --- Option (e) 160.0 mm

Answer:

(a) Magnetic moment will be [tex]17.212\times 10^{-4}A-m^2[/tex]

(b) Torque will be [tex]6.024\times 10^{-4}N-m[/tex]

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil [tex]A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2[/tex]

Current is given as [tex]i=15mA=15\times 10^{-3}A[/tex]

Number of turns N = 25

(A) We know that magnetic moment is given by [tex]magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2[/tex]

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by [tex]\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m[/tex]

Final answer:

a) The magnitude of the magnetic moment is 0.0172 A·m². b) The magnitude of the torque on the loop is 0 N·m.

Explanation:

a) To calculate the magnitude of the magnetic moment, we can use the formula μ = nIA, where μ is the magnetic moment, n is the number of turns, I is the current, and A is the area of the coil. In this case, n = 25, I = 15.0 mA = 0.015 A, and A = (5.40 cm)(8.50 cm) = 45.9 cm² = 0.00459 m². Plugging these values into the formula, we get μ = (25)(0.015 A)(0.00459 m²) = 0.0172 A·m².

b) To calculate the magnitude of the torque, we can use the formula τ = μBsinθ, where τ is the torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between μ and B. In this case, μ = 0.0172 A·m² (calculated in part a),

B = 0.350 T, and since the magnetic field is applied parallel to the plane of the loop, θ = 0°.

Plugging these values into the formula, we get τ = (0.0172 A·m²)(0.350 T)(sin 0°)

= 0 N·m.

Answer:

Explanation:

Let the two inclined planes having angle of inclinations α and β.

The acceleration along the inclined plane acting on the body is gSinα and gSinβ.

If α > β

So, g Sinα > g Sinβ

So, more the inclination of the plane more be the acceleration of body and hence the time taken is less.

So, the body kept on the inclined whose inclination is more reaches at the bottom first.

The mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.

The net horizontal force on the masses is calculated as follows;

[tex]\Sigma F_x = 0\\\\mgsin(\theta) = ma\\\\gsin(\theta) = a\ \ \ or \ \ \ \\\\gsin(\beta ) = a[/tex]

Assuming the value of the angles of inclination of the plane is the following;

a = gsin(60) = 0.866g

a = gsin(30) = 0.5g

Thus, we can conclude that the mass sitting on the plane inclined to a greater angle to the horizontal will reach the bottom first because it will experience greater acceleration.

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