At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?

Answer:[tex]\frac{5L}{6}[/tex]

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass [tex]M=32\ kg[/tex] is sitting at a distance of L from  center

if two child of mass [tex]\frac{M}{2}[/tex] is sitting at a distance [tex]\frac{L}{6}[/tex] and [tex]x[/tex](say) from pivot then net torque about pivot is zero

i.e.

[tex]\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx[/tex]

as [tex]\tau_{net}=0[/tex]

Therefore

[tex]MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx[/tex]

[tex]L-\frac{L}{6}=x[/tex]

[tex]x=\frac{5L}{6}[/tex]

Therefore another child is sitting at a distance of [tex]\frac{5L}{6}[/tex]

Answer:

The speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

[tex]v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s[/tex]

So, the speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].

The galaxy is receding from the Earth due to redshift. Using the redshift value, we can calculate the speed of the galaxy.

The discrepancy in the measured wavelengths of light from a distant galaxy compared to its original wavelength on Earth is indicative of the galaxy moving away from Earth. This phenomenon, known as redshift, occurs when an object is moving away from the observer, causing the wavelength of light to stretch.

To calculate the speed of the galaxy relative to Earth, we can use the equation v = zc, where v is the speed, z is the redshift, and c is the speed of light. By plugging in the given values of 438.6 nm and 434.1 nm for the measured and original wavelengths respectively, we can solve for z. Once we know z, we can calculate the speed of the galaxy.

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The decibel level of the remaining pigs is 34.02dB.

To solve this problem, we need to understand the relationship between the number of pigs and the intensity of sound. Since the intensity is proportional to the number of pigs, we can use the formula:

[tex]\[ \text{Intensity} = k \times \text{Number of pigs} \][/tex]

where ( k ) is a constant of proportionality.

Given that the noise level from 199 pigs is 74.3 dB, we can set up the equation:

[tex]\[ 74.3 = k \times 199 \][/tex]

First, let's solve for ( k ):

[tex]\[ k = \frac{74.3}{199} \][/tex]

Now, we can use this value of ( k ) to find the intensity when only 138 pigs remain:

[tex]\[ \text{Intensity}_{\text{new}} = k \times 138 \][/tex]

Finally, we can convert this intensity back into decibels using the formula:

[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(\text{Intensity}) \][/tex]

Now, let's calculate:

[tex]\[ k = \frac{74.3}{199} \][/tex]

[tex]\[ k \approx 0.3726 \][/tex]

[tex]\[ \text{Intensity}_{\text{new}} = 0.3726 \times 138 \][/tex]

[tex]\[ \text{Intensity}_{\text{new}} = 51.36 \][/tex]

[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(51.36) \][/tex]

[tex]\[ \text{Noise level (dB)} = 34.02 \][/tex]

So, the decibel level of the remaining pigs is 34.02dB.

Answer:5.1 meters

Explanation:

wavelength=3 x1.7

Wavelength=5.1 meters

Answer:[tex]W=16.837\ kJ[/tex]

Explanation:

Given

Drag force is given by

[tex]F_{drag}=A+Bv+Cv^2[/tex]

for [tex]A=220.5\ N[/tex]

[tex]B=-5.93\ N-s/m[/tex]

[tex]C=0.611N-s/m^2[/tex]

car accelerate from 0 to [tex]100\ km/hr[/tex] in [tex]3.5\ s[/tex]

so acceleration is given by

[tex]v=u+at[/tex]

here u=initial velocity is zero

[tex]v=100\km/hr\approx 27.78\ m/s[/tex]

[tex]27.78=0+a(3.5)[/tex]

[tex]a=7.936\ m/s^2[/tex]

Now work done is given by

[tex]dW=F\cdot vdt[/tex]

[tex]\int_{0}^{W}dW=\int_{0}^{3.5}F\cdot vdt[/tex]

[tex]W=\int_{0}^{3.5}[Av+Bv^2+Cv^3]dt[/tex]

[tex]W=\int_{0}^{3.5}[220.5at-5.93a^2t^2+0.611a^3t^3]dt[/tex]

[tex]W=\int_{0}^{3.5}220.5\times 7.936tdt-\int_{0}^{3.5}5.93\times (7.936)^2t^2dt+\int_{0}^{3.5}0.611\times (7.936)^3t^3dt[/tex]

[tex]W=1749.88[\frac{t^2}{2}]_0^{3.5}-373.47[\frac{t^3}{3}]_0^{3.5}+305.389[\frac{t^4}{4}]_0^{3.5}[/tex]

[tex]W=10,718.015-5337.508+11,456.85[/tex]

[tex]W=16.837\ kJ[/tex]

The work done by drag forces on a car accelerating from 0 to 100 km/hr can be calculated using the equation for total force and its given coefficients, by first converting the velocity to m/s, then finding the average velocity to use it to find the drag force, and finally using the net force and distance covered to calculate the work done.

The work done by the drag forces on a car accelerating steadily from 0 km/hr to 100 km/hr over 3.5 s can be calculated using the given equation for total force due to different effects, Fdrag=A+Bv+Cv2, and the coefficients for an Accord car. Please note that before calculating, the velocity needs to be converted from km/hr to m/s as SI units should be consistent.

First, find the average velocity of the car during its acceleration period, which is (0 + final velocity)/2. Then, substitute the values in the given equation Fdrag=A+Bv+Cv2 to find out the Net Force acting on the car.

Next, use this Net Force to find the Work done against the drag forces during the acceleration. The Work done (W) against a force is calculated by the equation W = F * d, where F is the net force (drag force) acting on the body and d is the total distance covered during the acceleration. To find the distance covered by the car, replace 's' in the equation s = ut + 1/2*a*t^2 with the average velocity of the car times the total time (t) it took to accelerate.

This will give you the total work done against drag forces on accelerating the car from 0 km/hr to 100 km/hr over 3.5 s.

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Answer:23.25 A

Explanation:

Given

Rating of toaster [tex]P_1=1060\ W[/tex]

Coffee pot [tex]P_2=500\ W[/tex]

microwave [tex]P_3=1230\ W[/tex]

Voltage applied [tex]V=120\ V[/tex]

if they are connected in parallel then all three operates at same voltage

so their resistance are

[tex]P=\frac{V^2}{R}[/tex]

thus [tex]R_1=\frac{V^2}{P_1}=\frac{120^2}{1060}[/tex]

[tex]R_1=13.58\ \Omega[/tex]

[tex]R_2=\frac{V^2}{P_2}=\frac{120^2}{500}[/tex]

[tex]R_2=28.8\ \Omega[/tex]

[tex]R_3=\frac{V^2}{P_3}=\frac{120^2}{1230}[/tex]

[tex]R_3=11.707\ \Omega[/tex]

and [tex]V=IR[/tex]

where I=current

thus [tex]I_1=\frac{V}{R_1}=\frac{120}{13.58}[/tex]

[tex]I_1=8.83\ A[/tex]

[tex]I_2=\frac{V}{R_2}=\frac{120}{28.8}[/tex]

[tex]I_2=4.16\ A[/tex]

[tex]I_3=\frac{V}{R_3}=\frac{120}{11.707}[/tex]

[tex]I_3=10.25\ A[/tex]

Total current [tex]I=I_1+I_2+I_3=23.24\ A[/tex]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle [tex]\theta _c = sin^{-1} (\frac{n}{n_{slab}} )[/tex]

[tex]sin \theta _ c =\frac{n}{n_{slab}}[/tex]

According to Snell's law of refraction:

[tex]n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)[/tex]

At point P ; [tex]90 - \theta _2 \leq \theta _c[/tex]

[tex]\theta _2 = 90 - \theta _c[/tex]

Therefore:

[tex]n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}[/tex]

Then maximum value of refractive index  n of the fluid is:

[tex]n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }[/tex]

[tex]n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }[/tex]

n = 1.4266

Answer:

a) 319.56 m/s

b) 0.8057 kg /s

c) 0.840 kg /s

Explanation:

Check the pictures attached for the whole explanation

Final answer:

The velocity of nitrogen at the nozzle exit and the mass flow rate depend on the conditions and properties of the gas, using principles such as the Bernoulli equation and the ideal gas law. Without specific constants, these values cannot be determined precisely. The maximum flow rate would be obtained by reducing the ambient pressure below the critical pressure ratio, causing choked flow.

Explanation:

To answer the student's question about the velocity of the nitrogen gas at the nozzle exit, we need to apply the principles of fluid dynamics. Under the provided conditions, without any losses and assuming adiabatic expansion, the velocity can be calculated using the Bernoulli equation and the equation of continuity. However, given that we only have the inlet pressure, outlet pressure, and temperature, we need additional equations like the ideal gas law to find the density at the exit. Unfortunately, without the specific heat ratio or gas constant for nitrogen, a precise calculation cannot be provided.

The mass flow rate can then be calculated by multiplying the outlet velocity by the outlet area and the density of nitrogen under the given conditions. An approximation can be made, but with the given data, it is not possible to provide an exact value since the outlet velocity is not known.

Regarding the maximum flow rate, it is related to the critical pressure ratio of the gas. If the ambient pressure is decreased below a certain critical point, the flow becomes choked, and the mass flow rate reaches a maximum. This maximum is a function of the chamber conditions and the specific properties of nitrogen.

The loop experiences a magnetic force to the left when leaving the field, a counterclockwise current flows when entering, a clockwise current flows when leaving, and a magnetic force to the left when entering the field.

The statement a is true. When leaving the magnetic field, the loop experiences a magnetic force to the left. This is because the magnetic field lines are directed from right to left in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.

The statement b is false. Upon entering the field, a counterclockwise current flows in the loop. This is because the increasing magnetic flux induces a current that opposes the increase, as stated by Lenz's law.

The statement c is true. Upon leaving the field, a clockwise current flows in the loop. This is because the decreasing magnetic flux induces a current that opposes the decrease, according to Lenz's law.

The statement d is false. When entering the field, the coil experiences a magnetic force to the left. This is because the magnetic field lines are directed from left to right in the field and the loop opposes the change in magnetic flux by generating a current that creates its own magnetic field, according to Lenz's law.

Answer:

The rotational inertia of my wheel is  [tex]I =1.083 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

      The mass of the wheel is [tex]m = 1.6 \ kg[/tex]

        The radius of the wheel is  [tex]r = 0.37 \ m[/tex]

        The height is  [tex]h = 6.7 m[/tex]

          The linear speed is  [tex]v = 4.7 m/s[/tex]

According to the law of energy conservation

             [tex]PE = KE + KE_R[/tex]

Where PE is the potential energy at the height h  which is mathematically represented as

             [tex]PE = mgh[/tex]

While KE is  the kinetic energy at the bottom of height h

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

Where  [tex]KE_R[/tex] is the rotational kinetic energy which is mathematically represented as

           [tex]KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]

Where  [tex]I[/tex] is the rotational inertia

       So  substituting this formula into the equation of energy conservation

         [tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]

=>       [tex]I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}[/tex]

substituting values

           [tex]I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}[/tex]

             [tex]I =1.083 \ kg \cdot m^2[/tex]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is [tex]\tau = 35.63\ N \cdot m[/tex]  

b

The distance of the single for equilibrium to occur is [tex]r_s =1.62 \ m[/tex]

Explanation:

From the question we are told that

     The mass of the left rock is  [tex]m_s = 2.25 \ kg[/tex]

     The mass of the rock on the right [tex]m_p = 10.1 kg[/tex]

    The distance from  fulcrum to the center of the pile of rocks is  [tex]r_p = 0.360 \ m[/tex]

Generally the torque produced by the pile of rock is mathematically represented as

           [tex]\tau = m_p * g * r_p[/tex]

Substituting values

         [tex]\tau = 10.1 * 9.8 * 0.360[/tex]                  

          [tex]\tau = 35.63\ N \cdot m[/tex]      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           [tex]\tau = m_s * g * r_s[/tex]

At equilibrium the both torque are equal

            [tex]35.63 = m_s * r_s * g[/tex]

Making [tex]r_s[/tex] the subject of the formula

             [tex]r_s = \frac{35.63 }{m_s * g}[/tex]

Substituting values

            [tex]r_s = \frac{35.63 }{2.25 * 9.8}[/tex]

            [tex]r_s =1.62 \ m[/tex]

Answer:

Normal force = 3N

Explanation:

We are given;

Coefficient of kinetic friction friction; μ_k = 0.3

Frictional force;F_f = 0.1N

Now,the formula for frictional force is;

F_f = μ_k*N

Where μ_k is coefficient of friction and N is the normal force.

So, making N the subject, we have;

N = μ_k/F_f

Plugging in the relevant values to obtain;

N = 0.3/0.1

N = 3N

Answer:

W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]

Explanation:

a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.

b) By the law of conservation of energy angular momentum

[tex]L_1=L_2[/tex]

[tex]I_1\omega_1=I_2\omega_2[/tex]

[tex]mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1[/tex]

[tex]\omega_2=(\frac{0.3}{0.15})^2\times2.85[/tex]

[tex]\omega_2=5.7\text{ rad/sec}[/tex]

c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)

[tex]K_1=\frac{1}{2} mr_1^2\omega_1^2[/tex]

[tex]K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2[/tex]

[tex]=9.12\times10^{-3} J[/tex]

Now,

[tex]K_2=\frac{1}{2} mr_2^2\omega_2^2[/tex]

[tex]K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2[/tex]

[tex]K_2=18.24\times10^{-3}[/tex] J

Therefore, Work done W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]

a. Angular momentum is conserved in this system. b. The change in kinetic energy of the block can be calculated. c. No work is done in pulling the cord.

a. Angular momentum is conserved when there are no external torques acting on the system. In this case, since the block is revolving on a frictionless surface, and there is no mention of any external torques, we can assume that angular momentum is conserved.

b. The change in kinetic energy of the block can be calculated using the equation ΔKE = KE_final - KE_initial. Since the block is modeled as a particle, its kinetic energy is given by KE = 1/2 * m * v^2, where m is the mass and v is the linear velocity. As the radius is changed, the linear velocity changes, and we can calculate the change in kinetic energy.

c. The work done in pulling the cord can be calculated using the equation W = ΔKE + ΔPE, where W is the work done, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy. In this case, since the block is on a frictionless surface, there is no change in potential energy, and we only need to calculate the change in kinetic energy.

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Answer:

water

Explanation:

Answer:water

Explanation:because it shaped the grand canyon

Answer:

louder

Explanation:

Since the two speakers producing same wavelength that are in phase,at the midpoint, the waves travel the same distance and hence path difference is zero

hence constructive interference takes place , due to this a louder sound is observed .

hence the answer is a) louder

When two speakers produce waves of the same wavelength that are in phase at a point midway between the speakers, constructive interference occurs. This results in the overlapping and combination of the waves to form a wave with higher amplitude, creating a louder sound.

The subject of this question involves the principle of wave interference in physics. This is a phenomenon that occurs when two waves come together while traveling through the same medium. At a point midway between the speakers, when two speakers produce waves of the same wavelength that are in phase, taking into account the path lengths traveled by the individual waves, you would expect to hear a louder sound.

This is a case of constructive interference, where the two sound waves, being in phase and of the same wavelength, will overlap and combine to form a wave with a greater amplitude, leading to a louder sound. This is explained in Figure 17.17 and 16.36, where the difference in the path lengths is one wavelength, resulting in total constructive interference and a resulting amplitude equal to twice the original amplitude.

However, it is worthy to note that in the real world recognition of this increased amplitude or louder sound will depend on the specific frequency of the sound, as sonic perception can vary with frequencies. This explanation is in reference to a single tone or frequency. When discussing music which is composed of many frequencies, the actual perception might be a bit more complex.

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Answer:36m/s

Explanation:

Distance=18m

time=0.5 seconds

speed=distance ➗ time

Speed=18 ➗ 0.5

Speed=36m/s

The question is about calculating the speed at which a baseball was thrown given that it traveled a distance of 18 meters in 0.5 seconds. By using the speed formula (speed = distance/time), we find that the baseball was thrown at a speed of 36 meters per second.

To calculate the speed of the baseball, we need to use the formula for speed which is speed = distance/time. Here, the distance covered by the baseball is 18m and the time taken is 0.5 seconds.

Substituting these values into the formula, we get speed = 18m / 0.5s = 36 m/s. So, the speed of the baseball is 36 meters per second.

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Answer:

Up And Down

Explanation:

In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.

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Final answer:

In a longitudinal wave on a horizontal spring, the medium oscillates back and forth horizontally, parallel to the direction of the wave's travel.

Explanation:

The question pertains to the behavior of longitudinal waves in a spring. When one student shakes the spring to create a longitudinal wave, the oscillation of the spring occurs in the same direction as the wave's propagation. This means that in a longitudinal wave, the medium—the spring in this case—oscillates parallel to the wave's direction of motion. Therefore, if the spring is held horizontally and the wave travels from left to right, the individual coils of the spring will move left and right along the same horizontal line, compressing and expanding as the wave passes through.

In contrast, with transverse waves, the medium moves perpendicularly to the direction of the wave's travel, such as in the motion of a rope being moved up and down while the wave travels horizontally.

Answer:

Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.

Explanation:

Answer:

m=0.893kg

Explanation:

time period of oscillations is given by= 2π√(m/k)

m: mass of the object

k: spring constant

when T=1.5 and m=0.415

1.5= 2π√(0.415/k)

k= 7.27 N/m

when T= 2.2s

2.2= 2π√(m/7.27)

m=0.893kg

The mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg

The formula for calculating the period of a simple pendulum is expressed as:

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

m is the mass of the spring

k is the spring constant

Given the following parameters

m = 0.415kg

T = 1.4 secs

Get the spring constant

[tex]1.4=2(3.14)\sqrt{\frac{0.415}{k} } \\1.4=6.28\sqrt{\frac{0.415}{k} }\\ 0.2229=\sqrt{\frac{0.415}{k} }\\ \frac{0.415}{k} =0.0497\\0.0497k=0.415\\k=\frac{0.415}{0.0497}\\k= 8.35N/m[/tex]

Given the period is 2.2secs, the mass of the spring will be expressed as:

[tex]2.2=2 (3.14)\sqrt{\frac{m}{8.35} } \\2.2=6.28\sqrt{\frac{m}{8.35} }\\\sqrt{\frac{m}{8.35} }=0.3503\\\frac{m}{8.35} =0.1227\\m=1.025kg[/tex]

Hence the mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg

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Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

Final answer:

The final velocity of the two football players after the collision is 0.738 m/s in the positive direction, calculated using the conservation of momentum.

Explanation:

The scenario described involves a conservation of momentum problem, where two football players collide and cling together. We can solve for the final velocity by using the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.

To find the final combined velocity of the players after the collision, we use the formula:

Momentum before collision = Momentum after collision

(m1 × v1) + (m2 × v2) = (m1 + m2) × v_final

Plugging in the given values:

(89.5 kg × 6.05 m/s) + (111 kg × (-3.55 m/s)) = (89.5 kg + 111 kg) × v_final

After calculating both sides,

541.475 kg·m/s - 393.55 kg·m/s = 200.5 kg · v_final

We get the final combined velocity,

v_final = 147.925 kg·m/s / 200.5 kg

v_final = 0.7377 m/s (to four significant figures)

The players will be moving with a velocity of 0.738 m/s in the positive direction.

Answer:

Option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.

Explanation:

According to Einstein's theory of relativity, the time between two observers, as one of them is moving respect to the other, is different, occurring the known time dilation. This theory indicates that, for the observer that is in an inertial frame of reference, he will measure a clock that is moving relative to him, to tick slower than his clock that is at rest. The faster the relative velocity, the greater the time dilation between the two of them.    

In our case, the person that is at rest (Albert) will measure the clock of the observer that is moving (Emmy) as ticking slower than his own watch, because Emmy is moving at high speed in relation to Albert (that is at rest), thus, time dilation is occurring for Emmy.                        

Therefore, the correct option is A, Albert observes that Emmy's watch is ticking more slowly than his watch.  

I hope it helps you!

Answer:

d) H0: Not all population means are equal,

HA: μ1 = μ2 = μ3 = μ4

Explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

Therefore, for the case above;

Since the underlying hypothesis is that average commute times are different across cities.

Null hypothesis H0: is that Not all population means are equal.

Alternative hypothesis Ha: μ1 = μ2 = μ3 = μ4

Answer:

Resistors in series in the circuit must always have the same current

Explanation:

Resistors are said to be connected in series if they are connected one after another.

The total resistance in the circuit with resistors connected in series is equal to the sum of individual resistances.

Individual resistors in series do not get the total source voltage. Total source voltage divide among them.

Answer:

it's a and d

Explanation:

because it all goes around if that makes sense.. and if u guys are here from apex

The answer to the questions are:

1. The resistivity for the material of the rod at 20 °C (ρ) is                                                

   [tex]1.26378\times 10^{-5} \Omega m[/tex]  .

2. The temperature coefficient of resistivity at 20 °C for the material of the              

    rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].

Given to us:

Voltage across the rod, V = 15.0 V

Length of rod, L = 1.5 m = 150 cm,

Diameter, d = 0.55 cm,

[tex]Radius, r= \frac{d}{2}=\frac{0.550}{2}=0.275\ cm[/tex]

[tex]\begin{aligned}Area, A&= \pi r^2\\&=\pi\times (0.275)^2\\&=0.075625\pi\ cm^2\\\end{aligned}[/tex]

Initial temperature,  [tex]T_o=20.0\ ^oC[/tex]

current at [tex]T_o[/tex],  [tex]I_o= 18.8\ A[/tex]

Final temperature,  [tex]T_1=92.0\ ^oC[/tex]

current at [tex]T_1[/tex],  [tex]I_1=17.4\ A[/tex]

1.) To find out the resistivity of the rod(ρ),

Resistant of the rod(R),

[tex]\begin{aligned}\\R_o&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{18.8} \\&=0.7979\ \Omega \\\end{aligned}[/tex]

Resistivity of the rod at [tex]20^o\ C[/tex](ρ),

[tex]\begin{aligned}\\\rho&=\frac{RA}{L}\\&=\frac{0.7979\times 0.075625\pi}{150}\\&=0.00126378\ \Omega cm\\&=1.26378\times 10^{-3} \Omega cm\\&=1.26378\times 10^{-5} \Omega m\\\end{aligned}[/tex]

Hence, the resistivity for the material of the rod at 20 °C (ρ) is [tex]1.26378\times 10^{-5} \Omega m[/tex]  .

2.) To find out the temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation,

[tex]{R_1} ={R_o}[ 1+\alpha(T_1-T_o)][/tex]

we need resistant of the rod([tex]R_1[/tex]),

[tex]\begin{aligned}\\R_1&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{17.4} \\&=0.862\ \Omega \\\end{aligned}[/tex]

Now, solving to get the value of α

[tex]\begin{aligned}{R_1} &={R_o}[ 1+\alpha(T_1-T_o)]\\0.862&=0.7979[1+\alpha(92-20)]\\\frac{0.862}{0.7979}&= [1+\alpha(72)]\\1.0804&=[1+\alpha(72)]\\0.0804&=\alpha(72)\\\alpha&=0.0011169\ ^oC^{-1}\\\alpha&=1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex]

Hence, the temperature coefficient of resistivity at 20 °C for the material of the rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].

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To find the resistivity and temperature coefficient of resistivity of the rod material, we need to use Ohm's Law and the formula for resistivity. The resistivity can be calculated using the resistance, area, and length of the rod, while the temperature coefficient of resistivity can be found by comparing resistivities at different temperatures.

To find the resistivity of the material of the rod at 20°C, we can use the formula:

Resistivity (ρ) = (Resistance × Area) / (Length)

First, we need to find the resistance of the rod using Ohm's Law:

Resistance (R) = Voltage (V) / Current (I)

Next, we can substitute the values into the formula to find the resistivity. The temperature coefficient of resistivity can be calculated using the equation:

α = (ρ₂ - ρ₁) / (ρ₁ × (T₂ - T₁))

Where ρ₁ and ρ₂ are the resistivities at temperatures T₁ and T₂ respectively.

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Answer:

Explanation:

length of vibration l = .6 m

mass per unit length m = 2 x 10⁻³ / .6

= 3.33 x 10⁻³ kg/ m

n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]

n is frequency of vibration , l is length , T is tension in the string .

Apply this formula in the first case

440 = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

Apply this formula for second case

n = [tex]\frac{1}{2\times.6} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

587 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{3.33\times10^{-3}} }[/tex]

Dividing

[tex]\frac{440}{587}[/tex] = [tex]\frac{l}{.6}[/tex]

l = .45 m .

The distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.

The given parameters;

The frequency of a sound wave in a stretched string is calculated as;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

[tex]f_1(2l_1) = f_2(2l_2)\\\\f_1l_1 = f_2l_2\\\\l_2 = \frac{f_1l_1}{f_2} \\\\l_2 = \frac{440 \times 0.6}{587} \\\\l_2 = 0.45 \ m\\\\l_2 = 45 \ cm[/tex]

Thus, the distance from the bridge where the cellist must put his finger to play the D5 note is 45 cm.

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Answer:

[tex]v_S=\sqrt{2}v_L[/tex]

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration [tex]a=\frac{v^2}{r}[/tex]. Since [tex]a_S=2a_L[/tex], we have that: [tex]\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}[/tex]

They take the same curve, so we have: [tex]r_S=r_L=R[/tex]

Which means: [tex]v_S^2=2v_L^2[/tex]

And finally we obtain: [tex]v_S=\sqrt{2}v_L[/tex]

Answer:

Power;P = 1372000 W

Explanation:

We are given;

Mass of person;m = 800 kg

Distance sprinted;h = 700m

Time taken;t = 4 seconds

Now, the formula for Power is;

Power = Work done/Time taken

Where work done = Force x Distance

Now, Force is expressed as;

F = mg

Where m is mass and g is acceleration due to gravity = 9.8 m/s²

Thus, plugging in the relevant values into power equation gives;

Power;P = 800 x 9.8 x 700/4

Power;P = 1372000 W

Answer:

Length of the rod is 0.043m

Explanation:

Mass of the bell (M) = 36kg

Distance of the center of mass from pivot = 0.55m

Moment of inertia (I) = 36.0kgm²

Mass of clipper (m) = 2.8kg

Length of the bell to ring = L

The period of the pendulum with small amplitude of oscillation is

T = 2π √(I / mgd)

Where g = acceleration due to gravity = 9.8 m/s²

T = 2π √(36 / 36*9.8*0.55)

T = √(0.1855)

T = 0.43s

The period of the pendulum is 0.43s

To find the length of the Clapper rod for which the bell ring slightly,

T = 2π√(L / g)

T² = 4π²L / g

T² *g = 4π²L

L = (T² * g) / 4π²

L = (9.8* 0.43²) / 4π²

L = 1.7287 / 39.478

L = 0.043m

The length of the Clapper rod for the bell to ring slightly is 0.043m

The magnitude of the induced emf in the coil, calculated using Faraday's Law of Electromagnetic Induction, is 6.18 V when a uniform magnetic field changes linearly over a time of 1.19 seconds.

To find the magnitude of the induced emf in the coil while the magnetic field is changing, we will apply Faraday's Law of Electromagnetic Induction which states that the emf induced in a circuit is proportional to the rate of change of the magnetic flux through the circuit. Given that the coil has 140 turns (N) and each side of the square frame is 35 cm (which we need to convert to meters to maintain SI units, 0.35 m), we can calculate the area (A) of the square frame as:

A = side x side = 0.35 m x 0.35 m = 0.1225 m2

The magnetic field (B) changes from 0 to -0.426 Wb/m2 over a time (t) of 1.19 s. The rate of change of magnetic field (dB/dt) is:

dB/dt = - ( 0 - (-0.426) ) / 1.19 s = 0.358 Wb/m2/s

Now using Faraday’s Law, the induced emf (ε) is given by:

ε = -N x (dΦ/dt) = -N x A x (dB/dt)

Plugging in the values we get:

ε = -140 x 0.1225 m2 x 0.358 Wb/m2/s = -6.18 V

The magnitude of the induced emf is therefore 6.18 V (taking the absolute value as the question asks for magnitude).