The simple pendulum above consists of a bob hanging from a light string. You wish to experimentally determine the frequency of the swinging pendulum. (a) By checking the line next to each appropriate item on the list below, select the equipment that you would need to do the experiment. ____ Meter Stick ____ Protractor ____ Additional string ____ Stopwatch ____ Photogate ____ Additional masses (b) Describe the experimental procedure that you would use. In your description, state the measurements you would make, how you would use the equipment to make them, and how you would determine the frequency from those measurements. (c) You next wish to discover which parameters of a pendulum affect its frequency. State one parameter that could be varied, describe how you would conduct the experiment, and indicate how you would analyze the data to show whether there is a dependence. (d) After swinging for a long time, the pendulum eventually comes to rest. Assume that the room is perfectly thermally insulated. How will the temperature of the room change while the pendulum comes to rest

Answer:

the coefficient of speed fluctuation is 0.08

the thickness of the rim is [tex]1.423*10^{-4}\ \ m[/tex]

Explanation:

a.Estimate the coefficient of speed fluctuation

Let first determine the average speed of the flywheel by using the expression:

[tex]n = \frac{n_1+n_2}{2}[/tex]

where;

[tex]n_1 =[/tex] minimum speed  = 240 rev/min

[tex]n_2 =[/tex] maximum speed = 260 rev/min

∴ [tex]n = \frac{240 +260}{2}[/tex]

n = 250 rev/ min

To find the coefficient of speed fluctuation; we have:

[tex]C_s = \frac{n_2-n_1}{n}[/tex]

[tex]C_s = \frac{260-240}{250}[/tex]

[tex]C_s = 0.08[/tex]

Hence; the coefficient of speed fluctuation is 0.08

b . If the weight of the spokes is neglected, determine the thickness of the rim

Let's start solving this process by finding the moment of inertia of the flywheel.

The moment of inertia of the fly wheel is given by the equation:

[tex]I = \frac {E_2-E_1}{C_s \omega^2}[/tex]

where ;

[tex]\omega = \frac{2 \pi *n}{60}[/tex]  (since we are converting to rad/s)

[tex]\omega = \frac{2 \pi *250}{60}[/tex]

= 26.18 rad/s

[tex]E_2-E_1[/tex] = the energy fluctuation = 6.75 J

[tex]I = \frac{6.75}{0.08*26.18^2}[/tex]

= 0.123 kg.m²

To determine the weight of the flywheel ; we have the following expression;

[tex]I = \frac{W}{8g}(D^2+d^2)[/tex]

[tex]W = \frac{8gl}{D^2_o+d_i^2}[/tex]

where;

[tex]D_0[/tex] = outer diameter = 1.5 m

[tex]d_i =[/tex] inner diameter = 1.4 m

[tex]W = \frac{8*9.81*0.123}{1.5^2+1.4^2}[/tex]

W = 2.29 N

Let employ the weight density of the cast-iron flywheel [tex]w \rho[/tex] = 70575.5N/m²

Then the volume of the flywheel:

[tex]V = \frac{W}{ w \ rho}[/tex]

[tex]V = \frac{2.29}{70575.5}[/tex]

[tex]V = 3.24*10^{-5} m^3[/tex]

Let t be the thickness of the rim;

the thickness of the rim can be calculate by using the formula;

[tex]V = \frac{\pi t}{4}(D^2-d^2)[/tex]

[tex]t = \frac{4V}{\pi(D^2-d^2)}[/tex]

[tex]t = \frac{4*3.24*10^{-5}}{\pi(1.5^2-1.4^2)}[/tex]

[tex]t = 1.423*10^{-4}m[/tex]

Hence, the thickness of the rim is [tex]1.423*10^{-4}\ \ m[/tex]

Answer:

A float switch

Explanation:

A float switch is a component of a wet-dry vacuum cleaner that is not present in a typical canister style residential cleaner​.

A float switch is a device used to detect the level of liquid within a wet-dry vacuum cleaner. The switch may be used as an indicator to control other devices.

Some float switches contain a two-stage switch. The associated pump is activated as the liquid rises to the trigger point of the first stage. The second stage will be triggered if the liquid continues to rises.

A wet-dry vacuum cleaner contains a water collection bin or moisture-resistant collection system, and a float mechanism to handle liquid spills, which are not present in typical canister-style residential vacuum cleaners.

The component of a wet-dry vacuum cleaner that is not present in a typical canister-style residential cleaner is specifically designed to handle liquid spills in addition to dry debris. This design typically includes a separate water collection bin or moisture-resistant collection system, which is absent in regular canister vacuums that are only suitable for dry pickup. Wet-dry vacuums also often contain a float mechanism that stops suction once the water reaches a certain level to prevent the motor from ingesting water.

Answer:

A. recycling allows resources to be reused, instead of dumped in landfills.

Explanation:

Answer: 0.3 m

So basically we want to think of this situation from the perspective of conservation of energy. As in, we can derive the velocity of both the swing and Stewie at the bottom of the swing via:

mgh = 0.5m*v^2

v = (2*g*h)^0.5 = (2*9.8*0.5)^0.5 = 3.13 m/s

This represents the velocity of Stewie and the swing at the bottom of the swing's path. Now, we will think of this from the perspective of conservation of momentum. Basically, the collective momentum of Stewie and the swing is equal to the sum of their subsequent momenta:

(5+2)*3.13 = 5*2+5*v(swing); v(swing) represents the velocity of the swing following Stewie jumping.

7*3.13 = 10 + 5v

v = 2.4 m/s

Now, we return to conservation of energy and find that the kinetic energy of the swing following Stewie jumping is equivalent to its final gravitational potential energy.

Based on the easily derivable formula: h=v^2/2g, we find that h = 2.4^2/19.6,

which is: 0.3 m.

If g is supposed to be 10 or you need a different degree of precision, then you can use my method. Hope this helps.

Answer:

Check the explanation

Explanation:

given

charge of [tex]1.11 * 10^{-10}[/tex] C

radius 250m

The potential at a point 1m

and 1 V

mass 2 micrograms

charge of +2 Coulombs

the potential is V = K Q / r

V = 9 X 109 X 1.11 X 10-10 / 1

V = 9 X 109 X 1.11 X 10-10

V = 0.999 volt

nearly it is V = 1 v

the energy stored is U = q X V

U = 2 X 1

U = 2 J

the energy is stored in this configuration is U = 2 J

When a star moves towards or away from Earth, its light is shifted to longer or shorter wavelengths, respectively. The red dwarf moving away from Earth and the blue dwarf moving away from Earth would have their light shifted towards longer wavelengths, resulting in a redshift.

The Doppler effect is the change in the observed frequency of sound or light waves due to the relative motion between the source of the waves and the observer.

When a star moves towards or away from Earth, its light is shifted to a longer or shorter wavelength, respectively.

In this case, the red dwarf moving away from Earth at 39.1 km/s would have its light shifted towards longer wavelengths, resulting in a redshift.

Similarly, the blue dwarf moving away from Earth at 25.9 km/s would also experience a redshift.

On the other hand, the red giant moving towards Earth at 23.3 km/s would have its light shifted towards shorter wavelengths, resulting in a blueshift.

The yellow dwarf moving transversely at 15.1 km/s would not exhibit any shift in its light because its motion is perpendicular to the line of sight.

The red dwarf moving transversely at 14.1 km/s would also not show any shift in its light.

Answer:

D)  E/2

Explanation:

i) To find the magnitude of electric field between the plates you use:

[tex]E=\frac{\Delta V}{d}[/tex]

where V is the voltage and d the separation between plates. If d is doubled you obtain:

[tex]E'=\frac{\Delta V}{2d}=\frac{1}{2}\frac{\Delta V}{d}=\frac{1}{2}E[/tex]

That is, the magnitude of the electric field is halved.

Then, the answer is D.

ii) The magnitude of E' is E/2 because the charge on the plates generates the field E.

Then, the answer is D again

Answer:

Explanation:

a ) If x be the position of n the bright fringe on the screen , following formula holds .

x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .

If we increase the value of λ or  wave length, x will increase so central fringe along with all the fringes will shift away from the centre .

If we increase the value of D or screen distance , it will also increase x ,  so fringes along with central fringe  will shift away from the center.

b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .

So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .

Answer:

0.99 T/s

Explanation:

Solution

From the example given, we recall that,

The emf induced in the loop is V = 7.0

Closed loop conductor with r = 1.5 m

∅B = BA

The emf =  d∅B/ dt

which is

d/dt  (BA)

so,

emf = A dB/ dt

emf =πr² dB/ dt

Now,

dB/dt = emf /πr²

=  7/π * (1.5)²

Therefore dB/dt = 0.99 T/s

Answer:[tex]3800\ W[/tex]

Explanation:

Given

Lawn mover running for [tex]t=20\ min [/tex]

and does [tex]W=4560\times 10^3\ J[/tex]

We know Power is rate of work i.e.

[tex]P=\frac{\text{Work}}{\text{time}}[/tex]

[tex]P=\frac{4560\times 10^3}{20\times 60}[/tex]

[tex]P=3800\ W[/tex]

Thus Power output is [tex]3800\ W[/tex]

Answer:

4.04m

Explanation:

First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:

[tex]U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}[/tex]

To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):

[tex]W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}[/tex]

where Ff is the friction force. By replacing the values of the parameters you obtain:

[tex]d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m[/tex]

hence, the distance to the original position is 3.96m+0.08m=4.04m

The 12 kg block travels a distance of 4 meters before coming to a stop. This is calculated using the principles of conservation of energy and work done by a force, in this case, the friction force.

The question involves a situation related to Physics, specifically kinematics and mechanics. In order to calculate how far the block travels, we first need to understand how the forces in this situation work. When the block is released, the potential energy stored in compressed spring converts into kinetic energy which propels the block forward. However, due to the presence of friction, this kinetic energy gradually diminishes causing the block to eventually come to a stop.

The first step is to find the initial speed of the block just when it is released. We can use the principle of conservation of energy for this: the spring potential energy is equal to the initial kinetic energy. For potential energy in a spring, we use the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the amount of compression (so PE = 0.5 * 10000 * 0.08^2 = 32 Joules). Kinetic energy is given by KE = 0.5 * m * v^2, where m is the mass and v is the velocity. Equating these (since initial PE = initial KE), we get v = sqrt((2 * PE) / m) = sqrt((2 * 32) / 12) = approx 4.08 m/s.

Now, knowing the initial speed, we can calculate how far the block travels before friction brings it to stop. The work done by the friction force will equal the initial kinetic energy of the block: Work = Friction force * distance = KE. Solving this equation for distance gives us distance = KE / Friction force = 32 Joules / 8 N = 4 meters.

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The total heat supplied to the gas in the process is calculated by considering the two expansion processes separately.

The total heat supplied to the gas in the process can be calculated by considering the two expansion processes separately.

In the first expansion, which is isobaric, the gas volume doubles. Since the pressure remains constant, the work done on the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume. The work done is equal to the heat supplied to the gas.

In the second expansion, which is adiabatic, the temperature returns to its initial value. In an adiabatic process, the work done on the gas is given by W = (γ - 1)ΔU, where γ is the ratio of specific heats and ΔU is the change in internal energy. Since the temperature returns to its initial value, the change in internal energy is zero, and therefore, no heat is supplied to the gas in this process.

Answer:

a) n2 = 1.55

b) 408.25 nm

c) 4.74*10^14 Hz

d) 1.93*10^8 m/s

Explanation:

a) To find the index of refraction of the syrup solution you use the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]   (1)

n1: index of refraction of air

n2: index of syrup solution

angle1: incidence angle

angle2: refraction angle

You replace the values of the parameter in (1) and calculate n2:

[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55[/tex]

b) To fond the wavelength in the solution you use:

[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm[/tex]

c) The frequency of the wave in the solution is:

[tex]v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz[/tex]

d) The speed in the solution is given by:

[tex]v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s[/tex]

Answer:

Choice A: approximately [tex]0.015\; \rm m^2[/tex], assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

[tex]\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].

By Pascal's Principle, because the first piston exerted a pressure of [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex] on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex]. In other words:

[tex]P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].

To achieve a force of [tex]9.900 \times 10^3\; \rm N[/tex], the surface area of the second piston should be:

[tex]\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}[/tex].

Answer:

a) -2.4*10^-4 V

   4.8*10^-5 A

b)  emf = 0

   induced current = 0

Explanation:

a) The induced emf is given by the following formula:

[tex]emf=-\frac{d\Phi_B}{dt}=-\frac{d(AB)}{dt}[/tex]    ( 1 )

A: area of the loop = 120 cm^2 = 120 (10^-2)^2 = 0.012m^2

The area is constant and dB/dt = 0.020T/s

By replacing in  the values of the parameters in the equation (1) you obtain:

[tex]emf=-A\frac{dB}{dt}=-(0.012m^2)(0.020T/s)=-2.4*10^{-4}V[/tex]

The induced current is:

[tex]I=\frac{emf}{R}=\frac{2.4*10^{-4}V}{5.0\Omega}=4.8*10^{-5}A[/tex]

b) If the loop is made of an insulator, electrons in the wire does not feel the change in the magnetic flux. Due to that, there is no a work over the electrons, and consequantly, there is neither emf nor induced current.

(a) The magnitude of the induced emf is 2.4 x 10⁻⁴ V.

(b) The induced current is 4.8 x 10⁻⁵ A.

The magnitude of the induced emf is determined by applying Faraday's law of electromagnetic induction.

emf = dΦ/dt

emf = BA/t

where;

emf = (0.02 x 0.012)

emf = 2.4 x 10⁻⁴ V

The induced current is calculated as follows;

emf = IR

I = emf/R

I = (2.4 x 10⁻⁴) / (5)

I = 4.8 x 10⁻⁵ A

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Final answer:

To calculate the number of loops for an inductor in a resonant circuit with a given capacitor and frequency, the resonant frequency formula can be used in conjunction with the inductance formula for a solenoid. These formulas involve calculations that may require additional information not provided in the question, such as the length of the solenoid.

Explanation:

Calculating the Required Number of Loops for the Inductor

To find the number of loops needed for a solenoid inductor to resonate with a 286 nF capacitor at 18.0 kHz, we must use the formula for the resonant frequency (f) of an LC circuit:

f = 1 / (2π√(LC))

Where:

C is the capacitance (286 nF or 286 × 10-9 F).

L is the inductance we need to calculate.

f is the resonant frequency (18.0 kHz or 18,000 Hz).

The inductance (L) of a solenoid inductor is given by:

L = (μ0× N2× A) / l

Where:

N is the number of loops.

A is the cross-sectional area of the coil (π × radius2).

l is the length of the solenoid.

μ0 is the permeability of free space (4π × 10-7 T·m/A).

By rearranging the resonant frequency formula to solve for L and then equating it to the inductance formula of a solenoid, we can determine the number of loops required for the solenoid to meet the resonance condition with the given capacitor at the specified frequency.

Electric flux through a cube with an external electric field is zero if there are no charges inside the cube, as Gauss's law states that the net flux is proportional to enclosed charge, leading to zero flux when the electric field vectors align perpendicularly with the area vectors of the cube's sides.

The electric flux through a cube can be analyzed when an electric field is applied. If the electric field is given by E' = ai + cj, where i and j are unit vectors along the x and y axes respectively, and there is no field component along b, zero flux is observed through most of the cube's surfaces due to the perpendicular directions of the electric field and the area vectors of the surfaces. Specifically, the sides of the cube have area vectors perpendicular to the given field components, resulting in a scalar product of zero. Consequently, the flux through these sides is zero. Gauss's law states that the net flux through a closed surface is proportional to the charge enclosed within the surface. If there are no charges inside the cube, the net flux must be zero regardless of whether the charges are outside the cube or if the cube has an induced electric field due to a changing current, as would be the case with a wire acting as an inductor passing through it.

Given Information:

Initial horizontal speed = Vx = 18 m/s

horizontal distance = R = 9 m

Required Information:

Vertical displacement = h = ?

Answer:

Vertical displacement = h = 1.23 m

Explanation:

The horizontal distance covered by the skater is given by

R = Vx*t

Where R is horizontal distance, Vx is the initial horizontal speed and t is the time taken for the jump.

t = R/Vx

t = 9/18

t = 0.5 seconds

The vertical displacement covered by the skater is given by

h = Vy*t + ½gt²

Where Vy is the initial vertical speed of the skater and is zero since the skater jumps horizontally, g is the gravitational acceleration and h is the vertical displacement.

h = 0*t + ½gt²

h = ½gt²

h = ½(9.81)(0.5)²

h = 1.23 m

Therefore, the vertical displacement of the skater is 1.23 m

Answer:

[tex]f_{police}=1268.7 Hz[/tex]    

Explanation:

We can use Doppler equation to find the frequency of the siren.

First of all we have the police car moving behind the car. Hence, the frequency detected by the car will be:  

[tex]f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})[/tex]      (1)

Now, when the police car is moving in front of the car, the frequency detected by the car will be:

[tex]f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})[/tex]        (2)            

By solving equation (1) and equation (2) for [tex]v_{police}[/tex] we have:

[tex]v_{police} = 44.67 m/s[/tex]

Knowing that:

Finally, we just need to put this value into the first equation to find frequency of the police car.

[tex]f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})[/tex]    

[tex]f_{police}=1310(\frac{343-44.7}{343-35})[/tex]  

[tex]f_{police}=1268.7 Hz[/tex]    

I hope it helps you!

Answer:

- Momentum is conserved

- Some kinetic energy is converted into thermal energy

Explanation:

Collision occurs when an object exert force on another. Collision can either be elastic or inelastic depending on whether the bodies stick together or separates after collision.

For elastic collision, the bodies separates after collision and due to this both their momentum and energy are conserved. Both separated object doesn't loose energy as such during collision and they possesses greater momentum.

In inelastic collision, the bodies stick together after collision. Their momentum is conserved but not their kinetic energy since they are not free to move independently.

Based on the question, the type of collision that occur is an inelastic collision since they stick together in air after collision. Some of the kinetic energy of the particles are turned into vibrational energy of the atoms leading to heating effect. Based on this, the following are true

- Their momentum before and after collision is conserved

- Some kinetic energy is converted into thermal energy(heat)

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

[tex]T_1+V_1=T_2+V_2[/tex]

[tex]0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}[/tex]

Also;

[tex]\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}[/tex]

Thus;

[tex]k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}[/tex]

where;

[tex]\delta[/tex] = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

[tex]k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}[/tex]

[tex]k = 104.82\ \ N/m[/tex]

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

[tex]T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0[/tex]

[tex]\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0[/tex]

[tex]\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0[/tex]

[tex]0.7344 \omega_3^2 = 2.128[/tex]

[tex]\omega _3 = \sqrt{\frac{2.128}{0.7344} }[/tex]

[tex]\omega _3 =1.70 \ rad/s[/tex]

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

A greater initial speed of the heavy cart could lead to a higher initial kinetic energy, which could result in a greater fraction of the kinetic energy being lost during a non-perfectly elastic collision. This could support the student's hypothesis.

The student's hypothesis could be correct because the fraction of kinetic energy lost from the two-cart system during the collision could indeed depend on the initial speed of the heavy cart. If the first, heavier cart is moving at a higher speed before the collision, it would have a higher initial kinetic energy (K init). When it collides with the second cart, more kinetic energy could be transformed into other forms of energy such as heat or sound, especially since the collision is not perfectly elastic. This means a greater fraction of kinetic energy could be lost (K lost) in the process, supporting the student's hypothesis.

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The higher initial speed of the first cart leads to greater deformation and energy conversion to non-mechanical forms, resulting in a larger fraction of kinetic energy loss may be the reason the hypothesis might be correct.

The hypothesis that a greater fraction of kinetic energy is lost from the system during the collision when the speed of the first cart is greater might be correct due to the nature of inelastic collisions and the role of deformation and heat generation during such collisions.

One reason for this is that the energy loss in a collision often includes contributions from factors like the deformation of the carts and the generation of heat, both of which can increase with the relative speed at the moment of collision. As the speed of the first cart increases, the impact forces during the collision are higher, leading to greater deformation and more energy being converted into internal energy (such as heat and sound) rather than being retained as kinetic energy in the system.

This means that when the speed of the first cart is higher, a larger portion of the initial kinetic energy is likely to be transformed into non-mechanical forms of energy, resulting in a greater fraction of kinetic energy being lost from the two-cart system.

Answer:

A) Since the confidence interval is not entirely above 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields.

Explanation:

See answer, it actually doubles as explaination.

Answer:

A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.

Explanation:

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.

Answer:

A. The resultant force in the same direction as the satellite’s acceleration.

Explanation:

Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.

The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.

In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.

In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.

The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.

Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.

Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.

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Answer: 12.8 A

Explanation:

Current is defined as the rate of flow of electric charge based on the formula:

I(current) = deltaQ(change in charge)/deltat(change in time).

First, however, we must convert the number of electrons into the number of coulombs. Based on the fact that the charge of 1 electron or 1 elementary charge is equal to 1.6*10^-19 C, we can calculate:

3.2*10^20 e = 1.6*10^-19*3*10^20 C = 51.2 C.

Now we use: I = Q/t = 51.2/4 = 12.8 A.

Hope this helped.

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

Final answer:

The physics problem involves two students on skateboards applying a force to each other and moving in opposite directions. To find the velocity of the second student, the conservation of momentum is used, considering no external forces act on them.

Explanation:

The subject of this question is physics, specifically relating to the principle of conservation of momentum in a system of two students standing on skateboards. When the two students push off each other, they start moving in opposite directions due to this principle. Conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this scenario, the system comprises the two students on skateboards.

To find how fast the second student moves, one would use the formula:

Total momentum before the shove = Total momentum after the shove

0 = (mass of student 1 \(\times\) velocity of student 1) + (mass of student 2 \(\times\) velocity of student 2)

We have the mass and velocity of student 1 (77 kg, moving left at 5.7 m/s), and the mass of student 2 (59 kg), but we need to calculate the velocity of student 2 after the push to ensure momentum is conserved.

Answer:

electric field due to the disk at the point x = 15 cm along the x-axis is;

E = 3.31 x 10^(9) N/C

Explanation:

We are given;

Radius;r = 30cm = 0.3m

Charge; Q = +3 μC = +3 x 10^(-6) C

Point, x = 15 cm = 0.15m

The formula for electric field due to the disk on the x-axis is given by;

E = [Q/(2ε₀•π•r²)] * [1 - (x/√(x² + r²))]

Where;

Q, x and r are as stated earlier

ε₀ is the permittivity of free space and has a constant value of 8.85 x 10⁻¹² C²/N.m

Thus, plugging in the relevant values, we have;

E = [3 x 10^(-6)/(8.85 x 10⁻¹² x π x 0.3²)] * [1 - (0.15/√(0.15² + 0.3²))]

E = 3.31 x 10^(9) N/C