A truck covers 45.0 m in 8.80 s while smoothly slowing down to final speed of 3.00 m/s. Find Its Original Speed.

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

[tex]\text{Initial Length} = L[/tex]

[tex]\text{Initial surface Area} = 6L^2[/tex] (Surface of a Cube)

[tex]\text{Initial Volume} = L^3[/tex]

The final measures

[tex]\text{Final Length} = L_f[/tex]

[tex]\text{Final surface area} = 6L_f^2[/tex]

[tex]\text{Final Volume} = L_f^3[/tex]

Given,

[tex]\frac{(SA)_f}{(SA)_i} = 2[/tex]

Now applying the same relation we have that

[tex](\frac{L_f}{L_i})^2 = 2[/tex]

[tex]\frac{L_f}{L_i} = \sqrt{2}[/tex]

The relation with volume would be

[tex]\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = 2.83[/tex]

Volume of the cube change by a factor of 2.83

Answer:

Explanation:

Given

Two block are connected by rope [tex]R_1[/tex]

[tex]R_2[/tex] rope is attached to block 2

suppose [tex]F_2[/tex] is a force applied to Rope [tex]R_2[/tex]

Applied force [tex]F_2[/tex]=Tension in Rope 2

[tex]F_2=(m_1+m_2)a---1[/tex]

where a=acceleration of system

Tension in rope [tex]R_1[/tex] is denoted by [tex]F_1[/tex]

[tex]F_1=m_1a---2[/tex]

divide 1 and 2 we get

[tex]\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}[/tex]

also [tex]m_1=2.11\cdot m_2[/tex]

[tex]\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}[/tex]

[tex]\frac{F_2}{F_1}=\frac{3.11}{2.11}[/tex]

[tex]\frac{F_1}{F_2}=\frac{2.11}{3.11}[/tex]

The ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.

Tension force:

When an object is connected with a rope such that the force exerted on the rope due to the weight of any object towards the upward direction, then such force is known as tension force.

Let R be the rope by which two blocks are connected. And R' is the rope attached to block 2.

Also, F' is the force applied to the rope R'. Then,

Applied force F' = Tension in the rope R'

[tex]F'=T\\\\ \dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}[/tex]

here, [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of ropes such that [tex]m_{1} = 2.11 \times m_{2}[/tex].

Solving as,

[tex]\dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}\\\\ \dfrac{F'}{T}= \dfrac{(2.11m_{2}+m_{2})a} {2.11 \times m_{2}a}\\\\\\ \dfrac{F'}{T}= \dfrac{3.11}{2.11} [/tex]

or

T/F' = 2.11 / 3.11

Thus, we can conclude that the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.

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Answer:

Time, t = 12 minutes

Explanation:

It is given that,

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

[tex]d=16-8+8-32+11.2=-4.8\ km[/tex]

d = 4800 m

Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s

Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{4800\ m}{6.66667\ m/s}[/tex]

t = 719.99 seconds

t = 720 seconds

or

t = 12 minutes

So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.                                      

To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be

[tex]W = PE[/tex]

[tex]W = mgh[/tex]

Here,

m = mass

g = Acceleration due to gravity

h = Height

our values are,

[tex]m = 0.345 kg[/tex]

[tex]g = 9.8m/s^2[/tex]

[tex]W = 111J[/tex]

Replacing,

[tex]W =mgh[/tex]

[tex]111 = (0.345)(9.8)h[/tex]

[tex]h = 32.83m[/tex]

Then the height is 32.83m.

Answer

given,

torque produced, τ = 22 N.m

Radius of the wheel. r = 10 cm

Moment of inertial = 2 kg.m²

initial angular speed = 0 rad/s

time, t = 5 s

a) we know,

   τ = I α

   22 = 2 x α

    α = 11 rad/s²

  using equation of rotation motion

[tex]\theta = \omega_o t + \dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta =\dfrac{1}{2}\times 11 \times 5^2[/tex]

  θ = 137.5 rad

  θ = 137.5/2π  =  22 revolution.

b) angular velocity of the motor

  [tex]\omega_f = \omega_i + \alpha t[/tex]

  [tex]\omega_f = 0 + 11 x 5[/tex]

  [tex]\omega_f = 55\ rad/s[/tex]

c) acceleration of a point on the rim of the wheel

  radial acceleration

 [tex]a_r = \omega^2 r[/tex]

 [tex]a_r = 55^2\times 0.1[/tex]

 [tex]a_r =302.5 \ m/s^2[/tex]

tangential acceleration of the point on the rim

  [tex]a_t = \alpha r[/tex]

  [tex]a_t = 11\times 0.1[/tex]

  [tex]a_t = 1.1\ m/s^2[/tex]

now, acceleration of the point

[tex]a = \sqrt{a_r^2+a_t^2}[/tex]

[tex]a = \sqrt{302.5^2+1.1^2}[/tex]

[tex]a = 302.5\ m/s^2[/tex]

Answer:

[tex]14m/s^2[/tex]

Explanation:

Initial speed of car=0 Km/h=[tex]0m/s[/tex]

Final speed of car=[tex]100km/h=100\times \frac{5}{18}=\frac{250}{9} m/s[/tex]

By using [tex]1km/h=\frac{5}{18}m/s[/tex]

Time taken by car=2 s

We know that

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

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Acceleration of car=[tex]\frac{\frac{250}{9}-0}{2}=\frac{250}{9\times 2}=13.9 m/s^2\approx 14m/s^2[/tex]

Hence, the acceleration of car about=[tex]14m/s^2[/tex]

Answer:

B) 3.50 m/s

Explanation:

The linear velocity in a circular motion is defined as:

[tex]v=\omega r(1)[/tex]

The angular frequency ([tex]\omega[/tex]) is defined as 2π times the frequency and r is the radius, that is, the distance from the center of the circular motion.

[tex]\omega=2\pi f(2)[/tex]

Replacing (2) in (1):

[tex]v=2\pi fr[/tex]

We have to convert the frequency to Hz:

[tex]7.18rpm*\frac{1Hz}{60rpm}=0.12Hz[/tex]

Finally, we calculate how fast is the child moving:

[tex]v=2\pi(0.12Hz)(4.65m)\\v=3.5\frac{m}{s}[/tex]

Answer:

1.58 W

Explanation:

Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is

[tex]A = \pi r^2 = \pi 12^2 = 452.389 m^2[/tex]

From the intensity of the sound we can calculate the power at 12 m

[tex]P = AI = 452.389 * 3.5\times10^{-3} = 1.58 W[/tex]

Josh, being 2 times heavier than Martin, reaches a height [tex]\( \frac{h}{16} \)[/tex] in a totally inelastic collision, making the correct answer (b).

In a totally inelastic collision, two objects stick together and move with a common final velocity. The conservation of linear momentum can be applied to find the final velocity and subsequently determine the height reached by the combined mass.

Let's denote the mass of Martin as [tex]\(m_M\)[/tex] and the mass of Josh as [tex]\(m_J\)[/tex], and let [tex]\(h_M\)[/tex] and [tex]\(h_J\)[/tex] be the heights they reach, respectively.

The conservation of linear momentum equation is given by:

[tex]\[ m_M \cdot v_M + m_J \cdot v_J = (m_M + m_J) \cdot v_f \][/tex]

where [tex]\(v_M\)[/tex] and [tex]\(v_J\)[/tex] are the initial velocities of Martin and Josh, and [tex]\(v_f\)[/tex] is their final velocity.

Since Martin and Josh stick together after the collision, their final velocity is the same [tex](\(v_f\))[/tex].

Now, given that Josh is 2 times heavier than Martin [tex](\(m_J = 2 \cdot m_M\))[/tex], we can express the initial momentum as:

[tex]\[ m_M \cdot v_M + (2 \cdot m_M) \cdot v_J = (3 \cdot m_M) \cdot v_f \][/tex]

Since the collision is totally inelastic, the final velocity is the same for both, and we can write:

[tex]\[ v_f = \frac{m_M \cdot v_M + 2 \cdot m_M \cdot v_J}{3 \cdot m_M} \][/tex]

Now, let's consider the conservation of energy. The potential energy lost during the collision is converted into kinetic energy, and finally, into potential energy again at the maximum height reached by the combined mass.

[tex]\[ m_M \cdot g \cdot h_M + m_J \cdot g \cdot h_J = \frac{1}{2} \cdot (m_M + m_J) \cdot v_f^2 \][/tex]

Substitute the expression for [tex]\(v_f\)[/tex] from the momentum equation into the energy equation:

[tex]\[ m_M \cdot g \cdot h_M + (2 \cdot m_M) \cdot g \cdot h_J = \frac{1}{2} \cdot (3 \cdot m_M) \cdot \left(\frac{m_M \cdot v_M + 2 \cdot m_M \cdot v_J}{3 \cdot m_M}\right)^2 \][/tex]

Simplify the equation and solve for [tex]\(h_J\)[/tex]. The final result should be [tex]\(h_J = \frac{h}{16}\)[/tex], so the correct answer is (b).

Answer:

The speed of the proton relative to the laboratory frame is 0.981c

Explanation:

Given that,

Speed of electron v= 0.90c

Speed of proton u= 0.70c

We need to calculate the speed of the proton relative to the laboratory frame

Using formula of speed

[tex]u'=\dfrac{u+v}{1+\dfrac{uv}{c^2}}[/tex]

Where, u = speed of the proton relative to the electron

v = speed of electron relative to the laboratory frame

Put the value into the formula

[tex]u'=\dfrac{0.70+0.90}{1+\dfrac{0.70\times0.90\times c^2}{c^2}}[/tex]

[tex]u'=c\dfrac{0.70+0.90}{1+(0.70\times0.90)}[/tex]

[tex]u'=0.981c[/tex]

Hence, The speed of the proton relative to the laboratory frame is 0.981c

Final answer:

The speed of the proton relative to the laboratory frame, when it moves to the right with a speed of 0.70c relative to an electron moving at 0.90c, is found to be approximately 0.98c using relativistic velocity addition.

Explanation:

To find the speed of the proton relative to the laboratory frame when an electron moves to the right with a speed of 0.90c and a proton moves to the right with a speed of 0.70c relative to the electron, we use the formula for relativistic velocity addition. This formula is given by:

V = (v + u) / (1 + vu/c²),

where V is the velocity of the proton relative to the lab, v is the velocity of the electron relative to the lab (0.90c), u is the velocity of the proton relative to the electron (0.70c), and c is the speed of light. Plugging in these values, we get:

V = (0.90c + 0.70c) / (1 + (0.90*0.70)) = 0.98c.

Therefore, the speed of the proton relative to the laboratory frame is approximately 0.98c, close to the speed of light, demonstrating the relativistic effects when velocities approach the speed of light.

Answer:

0.51 %

Explanation:

Since mass is the same at sea level and at 13000 m. And weight is the product of mass and gravitational acceleration g, the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level is essentially the percent reduction in the gravitational acceleration of an airplane cruising at 13,000 m relative to its gravitational acceleration at sea level, which is:

1 - 9.757 / 9.807 = 1 - 0.995 = 0.0051 = 0.51 %

Answer:

2.121876 mC

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

A = Area = [tex]0.037\times 12\ m^2[/tex]

d = Thickness = 0.0225 mm

E = Dielectric strength = [tex]1\times 10^8\ V/m[/tex]

k = Dielectric constant = 5.4

Capacitance is given by

[tex]C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F[/tex]

Maximum voltage is given by

[tex]V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V[/tex]

Maximum charge is given by

[tex]Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC[/tex]

The maximum charge that can be stored in this capacitor is 2.121876 mC

The maximum charge that can be stored in this capacitor is 16.016 μC.

To find the maximum charge that can be stored in the capacitor, we need to use the formula Q=CV, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance can be calculated using the formula C = (εrε0A)/d, where εr is the relative permittivity (dielectric constant) of mica, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

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Answer:

[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]

Explanation:

Given:

We know from the Coulomb's law:

[tex]F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}[/tex] ...........(1)

where:

[tex]\epsilon_0=[/tex] permittivity of free space

Also we have electric field due to Q at q (which is at a distance of 1 m):

[tex]E=\frac{1}{4\pi.\epsilon_0} \times \frac{Q}{r^2}\ [N.C^{-1}][/tex] ...........(2)

From (1) & (2)

[tex]E=\frac{F}{q}[/tex]

[tex]E=\frac{18}{10^{-9}}[/tex]

[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]

Answer:

magnitude of the electric field = 1.8 × [tex]10^{10}[/tex] N/C

Explanation:

given data

charge Q =2 C

charge q = 1 nC  = 1 × [tex]10^{-9}[/tex] C

distance r = 1 m

electric force = 18 N

solution

we know that here electric field due to Q is

E = k × [tex]\frac{Q}{r^2}[/tex]   ..............1

here q distance is 1 m

now we  apply here Coulomb’s law and get here electric force

F = k ×[tex]\frac{Q*q}{r^2}[/tex]   ..........2

we know constant k  = 8.988 × [tex]10^{9}[/tex] Nm²/C²

so from above both equation we get

electric filed = [tex]\frac{force}{charge}[/tex]    .................3

put here value

electric filed = [tex]\frac{18}{1*10^{-9}}[/tex] = 1.8 × [tex]10^{10}[/tex] N/C

To convert inches to centimeters, you multiply the number of inches by 2.54 (the conversion factor). Therefore 28.4 inches is approximately 72.1 centimeters.

In this physics problem, you are trying to change a distance measurement from inches to centimeters. Given the unit equality, 1 inch is equal to 2.54 centimeters. Therefore, to convert the measurement of 28.4 inches to centimeters, you need to multiply it by the conversion factor of 2.54 (i.e. 1 inch = 2.54 cm). So, you will simply multiply 28.4 inches by 2.54 to get the distance in centimeters: 28.4 inches * 2.54 cm/inch = approximately 72.1 centimeters.

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Answer: 171500 times faster.

Explanation: First let's take our Fluid medium to be air. The speed of sound in air is 343m/s or 343000mm/s.

Now, particle velocity is that which shows the alternating mean velocity of a particle in a medium(say in this case air).

A particle of air exposed to a standard sound pressure of 1 Pascal as a peak velocity amplitude of about 2mm/s.

Recall that the value of speed of sound in Air when converted to mm/s is 343000mm/s.

To get the number of times it's faster than the peak particle Velocity of air. We have to divide it by peak particle Velocity of air which is 2mm/s.

343000/2

=171500 times faster.

I believe this is clear.

The passengers in the car will hear a pitch of approximately 611.03 Hz.

The pitch that the passengers in the car will hear is determined by the Doppler effect. The frequency of a sound wave appears higher when the source is approaching and lower when it is moving away. The formula for calculating the observed frequency is f' = f((v + vd)/(v - vs)), where f' is the observed frequency, f is the source frequency, v is the speed of sound, vd is the velocity of the detector (passengers' car), and vs is the speed of the source (fire engine).

In this case, the frequency of the siren is 400 Hz, the speed of sound is 343 m/s, and the velocity of the detector car is 18 m/s. Since the source is moving south and the detector car is moving north, we take the velocities with opposite signs.

Using the formula f' = f((v + vd)/(v - vs)), we can calculate the observed frequency:

f' = 400((343 + 18)/(343 - (-35)))

f' ≈ 611.03 Hz

Therefore, the passengers in the car will hear a pitch of approximately 611.03 Hz.

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1) The final  angular speed is 4.85 rad/s

2) The change in kinetic energy is 1746 J

Explanation:

1)

The problem can be solved by applying the law of conservation of angular momentum: in fact, the total angular momentum of the system must be conserved,

[tex]L_1 = L_2[/tex]

The initial angular momentum is given by:

[tex]L_1 = (I_d + I_m) \omega[/tex]

where

[tex]I_d = \frac{1}{2}MR^2[/tex] is the moment of inertia of the cylinder, with

M = 63 kg is its mass

R = 2.5 m is the radius

[tex]I_m = mr^2[/tex] is the moment of inertia of the man, with

m = 97 kg is the mass of the man

r = 2.5 m is the distance of the man fro mthe axis of rotation

[tex]\omega=0.19 rev/s \cdot 2 \pi = 1.19 rad/s[/tex] is the angular speed

The final angular momentum is given by

[tex]L_2 = I_d \omega'[/tex]

where [tex]\omega'[/tex] is the final angular speed, and where the angular momentum of the man is zero because he is at the axis of rotation.

Combining the two equations, we get:

[tex](\frac{1}{2}MR^2 + mr^2) \omega = \frac{1}{2}MR^2 \omega'\\\omega'=(1+2\frac{m}{M})\omega=(1+2\frac{97}{63})(1.19)=4.85 rad/s[/tex]

2)

The initial kinetic energy (which is rotational kinetic energy) of the  system is given by

[tex]K_1 = \frac{1}{2}(I_d + I_m) \omega^2[/tex]

And substituting,

[tex]K_1 = \frac{1}{2}(\frac{1}{2}MR^2+mr^2)\omega^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2+(97)(2.5)^2)(1.19)^2=568.6 J[/tex]

The  final kinetic energy is given by

[tex]K_2 = \frac{1}{2}I_d \omega'^2[/tex]

And substituting,

[tex]K_2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2)(4.85)^2=2315 J[/tex]

So, the change in kinetic energy is

[tex]\Delta K = 2315 J - 568.6 J = 1746 J[/tex]

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The angular speed of the merry-go-round changes when the man moves closer to the center due to conservation of angular momentum, resulting in a new angular speed. The change in kinetic energy is calculated by comparing kinetic energy before and after the man's movement.

When a 97 kg man on a merry-go-round moves closer to the center, the angular speed of the system changes due to the conservation of angular momentum. The merry-go-round is described as a solid cylinder, and the movement of the man towards the center means his radius of rotation decreases, which leads to an increase in angular velocity. Initially, the man is 2.5 m away from the axis, standing on a merry-go-round that rotates at 0.19 rev/s. Upon reaching the center (0 m from the center), the entire system's angular momentum is conserved but concentrated within a smaller radius.

To calculate the new angular speed, we'll use the concept of moment of inertia (I) and the relation I1 * ω₁ = I2 * ω₂, where ω denotes angular speed. Since the man is no longer contributing to the rotational inertia when he moves to the center, only the merry-go-round's moment of inertia comes into play. Additionally, we'll need to convert the angular speed from rev/s to rad/s using the conversion factor: 1 rev = 2π rad.

The second part concerns the change in kinetic energy (KE) due to the man's movement. The rotational KE of a system is given by (1/2) * I * ω². By finding the KE before and after the man moves to the center, we can determine the change in kinetic energy.

Answer:

B=μ₀I/2r

Explanation:

Produced magnetic field due to an existing electric field through a coil or conductor can be explained by Biot-Savart Law. Formula for this law is:

dB=(μ₀I/4π.r²)dL

Here,

r=Radius of the loop

I and r are constants with respect to length L.

To convert linear displacement L into angular displacement Ф:

dL=r.dФ

So,

dB=(μ₀I/4π.r²)r.dФ

dB=(μ₀I/4π.r)dФ

Integrating both sides over the circle i.e. from 0 radians to 2π radians  (360⁰), while the integration will apply only on dФ as all others are constants.

B=(μ₀I/4πr)(2π-0)

B=(μ₀I/2r)

The magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].

The magnetic field produced due to current flowing through a coil given by Biot-Savart Law.

[tex]dB = \frac{\mu_o I}{4\pi r^2} dL[/tex]

where;

[tex]dB = \frac{\mu_o I}{4\pi r^2} .(rd \phi)\\\\dB = \frac{\mu_o I}{4\pi r} \ d \phi\\\\B = \frac{\mu_o I}{4\pi r} [\phi ]^{2\pi} _{0}\\\\B = \frac{\mu_o I}{4\pi r} (2\pi)\\\\B = \frac{\mu_o I}{2 r}[/tex]

Thus, the magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].

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Final answer:

To compute the ambient temperature, we can use the principle of heat transfer. By applying the formula Q = mcΔT, we can set up two equations to equate the heat lost by the coffee to the heat gained by the surroundings. Solving these equations will give us the ambient temperature.

Explanation:

To compute the ambient temperature, we can use the principle of heat transfer. The heat lost by the coffee is equal to the heat gained by the surroundings. We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

By applying this formula to the coffee, we can calculate the heat lost from t = 0 to t = 10 minutes and from t = 10 minutes to t = 20 minutes, and then equate it to the heat gained by the surroundings.

Let's assume the ambient temperature is T. Using the formula, we can set up the following equations:

By solving these equations, we can find the value of T, which is the ambient temperature.

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

[tex]V=\dfrac{W}{q}[/tex]

[tex]V=\dfrac{12}{0.0001}[/tex]

[tex]V=12\times 10^4\ Volt[/tex]

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

[tex]V=\dfrac{W}{q}[/tex]

[tex]V=\dfrac{24}{0.0002}[/tex]

[tex]V=12\times 10^4\ Volt[/tex]

Therefore, this is the required solution.

a) 4.95 s when the mass is halved

b) 7.00 s when the amplitude is doubled

c) 4.95 s when spring constant is doubled

d) 9.9 s when mass is doubled

Given:

T₀=7.0 s

The equation for the time period of the object is

[tex]T=2*\pi *\sqrt{\frac{m}{k} }[/tex]        ......................(i)

where m= mass of an object and k= spring constant

m= 1/2

∵T=7.0 s

On substituting value in equation (i)

Thus, T=4.95 s

Time period does not depend on amplitude thus,

T=7.0 s

Spring constant, k=2k

On substituting value in equation (i)

Thus, T=4.95 s

Mass=2m

On substituting value in equation (i)

Thus, T=9.9 s

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The period of a mass-spring oscillating system changes based on variations in mass and spring constant but not amplitude. If the mass is halved or the spring constant is doubled, the period decreases to approximately 4.95 s. If the mass is doubled, the period increases to approximately 9.90 s.

The period of oscillation for a mass attached to a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Keeping this formula in mind, let's evaluate how changes to the system will affect the period of oscillation:

Answer:

1.5T

Explanation:

Since, the amplitude of SHM is A. So, in one time period T the object will travel travel a distance of 4A.

6A-4A= 2A.

Now, this 2A distance must be traveled in T/2 time period.

So, the total time taken to travel a distance of 6A is T+T/2 = 3T/2 = 1.5T

In simple harmonic motion, an object oscillates between a maximum displacement (A) and its equilibrium point, completing a full cycle within a period (T). To move a total distance of 6A, it will need 1.5 cycles, thus taking 1.5T time.

The question you asked is about a simple harmonic oscillator, in this case, an object attached to a spring sliding on a frictionless surface. The motion of this object can be classified as simple harmonic motion (SHM).

In simple harmonic motion, the object oscillates between a maximum displacement or amplitude (A) and its equilibrium point. The time taken for the object to complete one full cycle, i.e., return to its starting point, is called the period (T).

To answer your question, when the object completes one full cycle of movement, it travels a distance of 4A (A to -A and -A to A). Therefore, if the object needs to travel a total distance of 6A, it will need 1.5 cycles. As the period, T, is the time taken for one cycle, the total time taken to travel 6A is 1.5T.

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Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

[tex]F = \sqrt{F_h^2+F_v^2}[/tex]

[tex]F = \sqrt{3.36^2+13.98^2}[/tex]

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

[tex]\theta =tan^{-1}(\dfrac{F_v}{F_h})[/tex]

[tex]\theta =tan^{-1}(\dfrac{13.98}{3.36})[/tex]

   θ = 76.48°

The resultant force will be 14.3844 N making an angle of 76.5° from the horizontal line.

Given to us

We know that a force is a vector quantity and can be divided into two component a vertical and a horizontal component. As shown below in the image.

The horizontal component is the cosine component while the vertical component is the sine component.

As the forces are pointing in different directions, therefore, the force on the left will be taken as positive while the force on the right is taken as negative.

[tex]F_{H} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{H} = [15\times Cos(45^o)]+[-8\times Cos(25^o)]\\F_H = 3.3561\ N[/tex]

As the forces are pointing in the same direction, therefore, the net force will be in the same direction,

[tex]F_{V} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{V} = [15\times Sin(45^o)]+[8\times Sin(25^o)]\\F_V = 13.9875\ N[/tex]

[tex]F_R = \sqrt{F_H^2+ F_V^2}[/tex]

[tex]F_R = \sqrt{3.3561^2+13.9875^2}\\F_R = 14.3844\ N[/tex]

[tex]Tan(\theta_R) = \dfrac{F_V}{F_H} = \dfrac{13.9875}{3.3561}= 4.167784\\\\(\theta_R) = Tan^{-1}(4.167784)\\(\theta_R) = 76.5^o[/tex]

As both the forces are positive the resultant force will be in the first quadrant, making an angle of 76.5° from the horizontal line.

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PART A) To give an approximation to the microscopic behavior of the drops, we must refer to Coulomb's law for which it is warned that similar charges repel. This spraying technique is used because when it starts to spray and expel the drops, they leave with the same loads and will be repelled, causing the layer of drops to be thinner and not thick drops. The lack of conglomeration of large drops will cause the paint to be finer.

PART B) When the object is loaded differently, even if the paint is shot in a different direction, it will end up adhered to the body by the action of this attraction

Charging paint droplets with an electric charge helps in achieving more even coverage and reducing overspray while painting a car.

a. When paint droplets are given a strong electric charge, they repel each other due to their like charges. This causes the droplets to spread out more evenly, resulting in a more uniform coverage on the car's surface.

b. By charging the droplets, they are attracted to the grounded surface of the car. This attraction helps ensure that most of the paint droplets move towards the car and stick to the surface, reducing the amount of overspray.

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Answer:

Here is the complete question.

A charge of 3 μC is at the origin. Sketch the function E x versus x for both positive and negative values of x. (Remember that Ex is negative when E points in the negative x direction.)

The graph is sketched in the attachment.

Explanation:

Since our electric field Eₓ = q/4πε₀x² (electric field for a point charge)

where q = electric charge = 3μC and x = distance from the origin of the charge. 1/4πε₀= 9 × 10⁹ Nm²/C².

Substituting the values into Eₓ  = 3 × 10⁻⁶×9 × 10⁹ /x²= 0.027/x². We plot values of x on the x- axis ranging from -3 to 3 into the equation to give us the graph for Eₓ. Note that the sign of Eₓ changes when x crosses the origin because of the direction of the electric field due to the new position of the charge. The graph is in the attachment below. Note that as x tends to zero, Eₓ tends to infinity and x tends to infinity, Eₓ tends to zero.

a) The independent variables are x and t

b) The parameters are [tex]k[/tex] (wave number), [tex]A[/tex] (the amplitude), [tex]\omega[/tex] (angular frequency)

c) The phase of this wave is zero

d) The wavelength of the wave is [tex]\lambda=\frac{2\pi}{k}[/tex]

e) The period of the wave is [tex]T=\frac{2\pi}{\omega}[/tex]

f) The speed of propagation of the wave is [tex]v=\frac{\omega}{k}[/tex]

Explanation:

a)

In physics and mathematics:

For the wave in this problem, therefore, we have:

b)

In physics and mathematics, the parameters of a function are the quantities whose value is constant (so, they do not change), and the value of the dependent variable also depends on the values of these parameters.

Therefore in this problem, for the function that represents the y-displacement of the wave, the parameters are all the constant factors in the formula that are not variables. Therefore, they are:

c)

The phase of this wave is zero.

In fact, a general equation for a wave is in the form

[tex]y(x,t)=Asin(kx-\omega t+\phi)[/tex]

where [tex]\phi[/tex] is the phase of the wave, and it represents the initial angular displacement of the wave when x = 0 and t = 0.

However, the equation of the wave in this problem is

[tex]y(x,t)=Asin(kx-\omega t)[/tex]

Therefore, we see that its phase is zero:

[tex]\phi=0[/tex]

d)

The wavelength of a wave is related to the wave number by the following equation

[tex]k=\frac{2\pi}{\lambda}[/tex]

where

k is the wave number

[tex]\lambda[/tex] is the wavelength

For the wave in this problem, we know its wave number, [tex]k[/tex], therefore we can find its wavelength by re-arranging the equation above:

[tex]\lambda=\frac{2\pi}{k}[/tex]

e)

The period of a wave is related to its angular frequency by the following equation

[tex]\omega=\frac{2\pi}{T}[/tex]

where

[tex]\omega[/tex] is the angular frequency

T is the period of the wave

For the wave in this problem, we know its angular frequency [tex]\omega[/tex], therefore we can find its period by re-arranging the equation above:

[tex]T=\frac{2\pi}{\omega}[/tex]

f)

The speed of propagation of a wave is given by the so-called wave equation:

[tex]v=f\lambda[/tex]

where

v is the speed of propagation of the wave

f is the frequency

[tex]\lambda[/tex] is the wavelength

The frequency is related to the period of the wave by

[tex]f=\frac{1}{T}[/tex]

So, we can rewrite the wave equation as

[tex]v=\frac{\lambda}{T}[/tex]

From part d) and e), we found an expression for both the wavelength and the period:

[tex]\lambda=\frac{2\pi}{k}\\T=\frac{2\pi}{\omega}[/tex]

Therefore, we can rewrite the speed of the wave as:

[tex]v=\frac{2\pi/k}{2\pi/\omega}=\frac{\omega}{k}[/tex]

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Answer:

a) v = 44.72 m/s

b) temperature rise = 2.22 K

Explanation:

A)

Data provided in the question:

Mass of the iron block = 1 kg

Kinetic Energy supplied = 1 KJ = 1000 J

Now,

Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]

here,

v is the velocity

thus,

1000 = [tex]\frac{1}{2}\times1\times v^2[/tex]

or

v² = 2000

or

v = 44.72 m/s

b)

Heat capacity of iron = 25.10 J/(mol K)

Molecular weight of iron = 55.85

Energy supplied = 1 kJ = 1000 J

Now,

Energy supplied = Mass × C × Change in temperature

Here, C = Heat capacity of iron ÷ Molecular weight of iron

= 25.10 ÷ 55.85

= 0.45 J/(g.K)

thus,

1000 = 1 kg × 0.45 × Change in temperature

or

1000 = 1000 g × 0.45 × Change in temperature

or

Change in temperature i.e temperature rise = 2.22 K

Answer: a. E =9.9*EXP(-19)J

1 mole E= 596178J

b. E= 1.32*EXP(-15)J, 1 mole E=795MegaJ

c. E= 1.98*EXP(-23)J

1 mole E = 11.9J

Explanation: The Energy of a photon E, the wavelength are related by

E= h*c/wavelength

h is the Planck's constant 6.6*EXP(-34)J.s

c is speed of light 3*EXP(8)m/s

h*c=1.98*EXP(-25)

Now let's solve

a. E = h*c/wavelength

= h*c/(200*EXP(-9)m

=9.9*EXP(-19)J

1 mole of a photon contian 6.022*EXP(23)photons by advogadro

Now to get the energy of 1 mole of the photon we have

9.9*EXP(-19)*6.023*EXP(23)

=596178J

b. E=h*c/150*EXP(-12)m

=1.32*EXP(-15)J

1 mole will have

1.32*EXP(-15)*6.022*EXP(23)J

=795*EXP(6)J

c. E= h*c/1*EXP(-2)m

=1.98*EXP(-23)J

1 mole of the photon will have

1.98*EXP(-23)J *6.022*EXP(23)

= 11.9J.

You will notice that the longer the wavelength of the photon the lesser the Energy it as.

NOTE: EXP represent 10^

Final answer:

The energy per photon and per mole for ultraviolet radiation at 200 nm, X-ray radiation at 150 pm, and microwave radiation at 1.00 cm can be calculated using Planck's equation, with the resulting energies being 9.939 x 10^-19 J, 1.327 x 10^-15 J, and 1.987 x 10^-23 J respectively, and the energies per mole being 5.98 x 10^5 J/mole, 7.99 x 10^8 J/mole, and 1.20 J/mole respectively.

Explanation:

To calculate the energy per photon and the energy per mole of photons for radiation of various wavelengths, we can use the relationship given by Planck's equation, which relates the energy (E) of a photon to its wavelength (λ):

E = hc / λ

where h is Planck's constant (6.626 x 10-34 J•s), c is the speed of light in a vacuum (3.00 x 108 m/s), and λ is the wavelength of the radiation. To find the energy per mole of photons, we can multiply the energy per photon by Avogadro's number (6.022 x 1023 photons/mole). Let's calculate the energy for each type of radiation:

To get the energy per mole, multiply each of these values by Avogadro's number:

Answer:

95.78125%

18000 m/s

22233.33 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v=u+at\\\Rightarrow v=0+20\times 15\times 60\\\Rightarrow v=18000\ m/s[/tex]

The velocity of the rocket at the end of the first 15 minutes is 18000 m/s which is the maximum speed of the rocket in the complete journey.

Distance traveled while speeding up

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 20\times 900^2\\\Rightarrow s=8100000\ m[/tex]

Distance traveled while slowing down

[tex]s=18000\times 900+\dfrac{1}{2}\times -20\times 900^2\\\Rightarrow s=8100000\ m[/tex]

Distance traveled during constant speed

[tex]384000000-(2\times 8100000)=367800000\ m[/tex]

Fraction

[tex]\dfrac{367800000}{384000000}\times 100=95.78125\ \%[/tex]

Fraction of the total distance is traveled at constant speed is 95.78125%

Time taken at constant speed

[tex]t=\dfrac{367800000}{18000}=20433.33\ s[/tex]

Total time taken is [tex]900+20433.33+900=22233.33\ s[/tex]

The maximum velocity of the spaceship is 18000 m/s. The spaceship travels 95.78% of the total distance at this constant speed. The total time required for the trip is 370 minutes.

To answer the first part of the question, we can use the physics formula for velocity, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. As the spaceship starts from rest, u = 0. The acceleration a is given as 20 m/s² and the time t for which the spaceship is accelerating is 15min which is equal to 900 seconds. So, the maximum velocity, v = 0 + 20*900 = 18000 m/s.

For the second part, we understand that the spaceship spends an equal amount of time accelerating and decelerating, and the rest of the time it travels at a constant velocity. Therefore, during acceleration and deceleration, it follows a distance d = 1/2 * a * t², with a = 20 m/s² and t = 900 s. We find the distance covered during acceleration and deceleration, d = 1/2 * 20 * 900² = 8100000 m. Given the total distance from Earth to Moon is 384000000 m, the fraction of the distance at a constant speed is 1 - (2*8100000)/384000000 = 0.9578 or 95.78%.

Finally, to find the total time of the trip, we know that the time spent accelerating and decelerating is 15min + 15min = 30min. The distance covered at constant speed is 0.9578 * 384000000m = 367680000m, and the speed is 18000 m/s. So, the time at constant speed is 367680000 / 18000 = 20400s = 340 min. Therefore, the total time of the trip is 30min + 340min = 370 min.

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Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

Final answer:

The highest speed in Michael Johnson's world-record 400 m sprint was achieved during the segment between the 100 m mark and the 200 m mark with a speed of 9.88 m/s.

Explanation:

To determine which 100 m segment Michael Johnson had the highest speed, we should calculate his speed for each segment separately. Speed is calculated by dividing the distance by the time taken to cover that distance. We are given the total time at each 100 m interval, so we need to calculate the time for each interval separately and then calculate the speed.

Comparing these speeds, we can see that the highest speed was achieved during the segment between the 100 m mark and the 200 m mark.