A ball of mass 24 g is attached to a cord of length 0.463 m and rotates in a vertical circle. What is the minimum speed the ball must have at the top of the circle so the cord does not become slack? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Neither Technician A nor B is correct.

Answer: Option D

Explanation:

In AC generator, diodes are usually used to rectify the alternator current. This is the transformation into direct current from alternating current. Sinusoidal voltage is generated by generators when it rotates in both directions. The output voltage [tex]V_{i n}[/tex] is converted to DC with diode help.

This process is called rectification. The output voltage [tex]V_{i n}[/tex]  is treated as the input signal of the diode circuit. So, after rectification, the sine wave is not produced by an AC generator. And, the sine wave resulted by AC generator when the output passes via diodes and not a straight line.

The waveform produced by an AC generator after rectification is a pulsating DC waveform, not a sine wave. Technician A is correct in this case.

Technician A is correct. The waveform produced by an AC generator after rectification is not a sine wave, but a pulsating DC waveform. After the output of the generator moves through the diodes, the negative half-cycles are clipped, resulting in a waveform that resembles a series of straight lines instead of a smooth sine wave. This pulsating DC waveform still contains alternating current components, but it is not a constant DC voltage.

Answer:

Explanation:

Speed: Speed can be defined as the ratio of distance to time. The S. I unit of speed is m/s. And it is expressed mathematically as,

speed = distance/time

S = d/t............................. Equation 1

Conversion: (i) If 1 miles = 1609.344 m,

                 then, 26.2 miles = 1609.344× 26.2

= 42164.813 m.

                    (i) if  1 hours = 3600 seconds,

               then,   5.5 hours = 5.5×3600

=19800 seconds.

Given: d = 26.7 miles= 42164.813 m, t = 5.5 hours = 19800 seconds

Substituting these values into equation 1

S = 42164.813/19800

S = 2.1 m/s.

Therefore Kenneth's average speed = 2.1 m/s

Answer:4.8 mph on edge

Explanation: have a great day :)

Answer:

Option d

Explanation:

Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.

Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.  

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice.

Answer: Option C

Explanation:

The closure temperature can also be termed as blocking temperature. It is mostly used in radiometric dating. As the temperature decreases, below a certain point the isotopes may get freeze in their lattice positions. And there may be slowing of diffusion.

At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.

Final answer:

Closure temperature is the point where isotopes are trapped within a crystal lattice, marking the mineral as a closed system, which is crucial for isotopic dating.

Explanation:

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice. This is a point in the cooling process of a rock or mineral, after crystallization, at which its isotopic constituents are no longer able to escape from its crystal lattice. Essentially, it is the temperature below which a mineral becomes a closed system for certain isotopes. Until reaching this temperature, isotopic parent and daughter products can freely exchange with the environment. As the temperature drops below the closure temperature, the system locks in place, preserving the isotopic date that will be analyzed to determine the history of the rock.

The speed of the block after the bullet exits is 7.5 m/s. This velocity is found by using the conservation of momentum principle and plugging in the known values into the momentum equation to solve for the unknown variable.

To solve for the speed of the wooden block after the bullet exits, we will use the principle of conservation of momentum. This principle states that in an isolated system, the total momentum before an event is equal to the total momentum after the event. The formula to calculate the final velocity of the block is:

[tex]m_{bullet} \times u_{bullet} + m_{block} \times u_{block} = m_{bullet} \times v_{bullet} + m_{block} \times v_{block}[/tex]

Where:

[tex]m_{bullet}[/tex] = mass of the bullet

[tex]u_{bullet}[/tex] = initial velocity of the bullet

[tex]m_{block}[/tex] = mass of the block

[tex]u_{block}[/tex] = initial velocity of the block (0 m/s, since it's stationary)

[tex]v_{bullet}[/tex] = final velocity of the bullet after passing through the block

[tex]v_{block}[/tex] = final velocity of the block

Given:

[tex]m_{bullet}[/tex] = 0.050 kg (50 g)

[tex]u_{bullet}[/tex] = 500 m/s

[tex]m_{block}[/tex] = 2 kg

[tex]u_{block}[/tex] = 0 m/s

[tex]v_{bullet}[/tex] = 200 m/s

Plugging in the values:

0.050 × 500 + 2 × 0 = 0.050 × 200 + 2 × [tex]v_{block}[/tex]

Solving for [tex]v_{block}[/tex]:

[tex]v_{block}[/tex] = (0.050 × 500 - 0.050 × 200) / 2

[tex]v_{block}[/tex] = (25 - 10) / 2

[tex]v_{block}[/tex] = 15 / 2

[tex]v_{block}[/tex] = 7.5 m/s

Therefore, the velocity of the block after the bullet exits is 7.5 m/s.

Answer:

ΔH = -57.78kj/mol

Explanation:

Assumptions

These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/goC.

  the heat  evolved will be

q = mcΔt

first of all convert ml to grams

(43.6 mL + 43.6 mL ) = 87.2mL of solution.

density of water=1g/ml

87.2 mL X 1     g/ml        =  87.2 grams of solution.

(mass = Volume X Density)

Find the temperature change.

      Δt =tfinal - tinitial = 26.88oC - 19.95oC = 6.93oC

   q = mcΔt

      = 87.2 grams X 4.184 J/goC    X 6.9oC

      = 2.52 X 10^3 J

This is the heat gained by the water, but in fact it is the heat lost by the reacting HBr and NaOH, therefore q = -2.52 x 10^3 J.

 i.e. it is an exothermic reaction, heat was lost to the water and it got warmer

to find the how much of HBr that was used in mol

43.6 mL of HBr X 1.00 mol HBr/ 1000 mL HBr = 0.0436 mol HBr

same quantity of base NaoH was used

. molar enthalpy = J/mol = -2.52 x 10^3 J /  0.0436 mol

-57.78kj/mol

Therefore, for the neutralization of HBrl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -57.78kj/mol

Explanation:

Mass, m = 899 kg

Initial velocity, u = 0 m/s

Final velocity, v = 26.3 m/s

Time, t = 0.59 s

a) We have equation of motion v = u + at

   Substituting

                     v = u + at

                     26.3 = 0 + a x 0.59

                       a = 44.58 m/s²

     Acceleration of dragster = 44.58 m/s²

b) Force = Mass x Acceleration

   F = ma

   F = 899 x 44.58 = 40074.07 N

   F = 40.07 kN

  Average net force on the dragster during this time is 40.07 kN

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = [tex]\frac{W}{t}[/tex]

Substitution

[tex]P = \frac{5}{3.7}[/tex]

Result

P = 1.35 W

Answer: No, because the atoms are arranged differently. Looking at 100 atoms in each sample, the volume would be smaller in a diamond, because the atoms are packed more closely together. The density of a diamond would be higher.

Explanation:

Final answer:

Graphite and diamond have different densities due to the varied arrangements of carbon atoms in their structures, with diamonds being denser and harder due to a three-dimensional bond network.

Explanation:

The difference in density between graphite and diamond can be attributed to the variation in how carbon atoms are arranged in these two allotropes of carbon. Graphite has a layered structure with weak bonds between each layer, allowing the sheets to easily slip past each other, leading to its softer structure and lower density.

In contrast, diamonds have carbon atoms bonded together in all three dimensions, creating a dense, strong lattice, which is why diamonds are very hard and have a higher density compared to graphite.

Answer:

10 mph

Explanation:

The above consequences double for every 10 mph over

50 mph that a vehicle travels.

Answer:

10 mph

Explanation:

Exceeding speed limit is one of the major causes of road accident.

Crash severity increases with the speed of the vehicle at impact.

The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.

The above consequences double for every 10 mph over

50 mph that a vehicle travels

Explanation:

Answer:braking

Explanation:

Large vehicles as compared to small vehicles require long braking distance, otherwise, it could topple the heavy vehicles. Heavy vehicles provide high torque thus it is used to carry heavy loads.

They run at relatively low speed as compared to the light vehicles as they are slow to accelerate and thus require long braking distance as Momentum associated with them is very high.

If sudden brakes are applied it may cause the vehicle to skid and flip over it due to the presence of large momentum.

The speed of the wave is 208 m/s

Explanation:

The wave equation states that ; speed of wave= wavelength *frequency

Mathematically, v=λ*f

Given;

λ=52 cm= 52/100 =0.52 m

f= 400 Hz

v=?

v=0.52*400 =208 m/s

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Answer:

Explanation:

Given

[tex]F(x)=x^3[/tex]

Rate of change of F(x) is given by

[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}[/tex]

[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}[/tex]

[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}[/tex]

[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}[/tex]

[tex]F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh[/tex]

Putting limits

[tex]F'(x)=3x^2[/tex]

The average rate of change will be "3x²".

According to the question,

The function, f(x) = x³

then,

f(x + h) = (x + h)³

Now,

→ f(x + h) - f(x) = (x + h)³ - x³

                      = x³ + h³ + 3x²h + 3xh² - x³

                      = h³ + 3x²h + 3xh²

and,

→ [tex]\frac{f(x+h) -f(x)}{h}[/tex] = [tex]\frac{h^3+3x^2h+3xh^2}{h}[/tex]

                    = [tex]\frac{h[h^2+3x^2+3xh]}{h}[/tex]

                    = h² + 3x² + 3xh

By applying the limit, we get

→ [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex] = [tex]\lim_{h \to 0}[/tex] h² + 3x² + 3xh

By substituting the values,

                                = 0² + 3x² + 3x(0)

                                = 3x²

Thus the above approach is correct.      

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Answer:

409.87803 m/s

Explanation:

v = Velocity of bullet

L = Latent heat of fusion = 6 cal/g

c = Specific heat of lead = 0.03 cal/g°C

[tex]\Delta T[/tex] = Change in temperature = (327-27)

m = Mass of bullet

[tex]1\ J=4.2\ J/cal[/tex]

The heat will be given by the kinetic energy of the bullet

[tex]Q=\dfrac{1}{2}mv^2[/tex]

According to the question

[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]

This heat will balance the heat going into the obstacle

[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]

The speed of the bullet is 409.87803 m/s

Answer:

12 A

Explanation:

Voltage, V = 120 V

Resistance, R  10 ohm

By using ohm's law

V = i x R

where, i is the current

i = V / R

i = 120 / 10

i = 12 A

thus, the current is 12 A.

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

Answer:

468449163762.0812 W

Explanation:

m = Mass = [tex]\rhoV[/tex]

V = Volume =[tex]\dfrac{4}{3}\pi r^3[/tex]

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

[tex]\rho[/tex] = Density = 19 g/cm³

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

A = Area = [tex]\pi R^2[/tex]

I = Intensity = [tex]\dfrac{P}{4\pi r^2}[/tex]

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

[tex]F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}[/tex]

According to the question

[tex]F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W[/tex]

The power required of the light source is 468449163762.0812 W

Final answer:

To calculate the power required of the light source, you need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The power is given by P = I * A.

Explanation:

To calculate the power required of the light source, we first need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The radiation force is given by the formula:

Fr = (P / c)A

where Fr is the radiation force, P is the power of the light source, c is the speed of light, and A is the cross-sectional area of the sphere.

Since the sphere is totally absorbing, all the light incident on it is absorbed, so the intensity of the light is given by:

I = P / A

where I is the intensity of the light. Rearranging the equation, we get:

P = I * A

Substituting the value of A for the area of the sphere, we can find the power required.

Answer: Astronauts only float around in the shuttle when they are outside the gravitational pull of the earth

Explanation: when astronauts takes off from the earth, they get to a point (space) where the earth's gravity can no longer pull them. At this state, they experience weightlessness because there is no gravity. Since there is no gravity to pull them down, hence they start floating.

Astronauts and objects float in the shuttle due to the microgravity state from continual free-fall around the Earth, creating the feeling of weightlessness.

Astronauts and objects like cans of soda float in the shuttle because of the lack of gravity in space, a state known as microgravity. When the shuttle is orbiting the earth, it's actually falling towards the earth but also moving forward. This forward motion allows the shuttle and everything in it to keep missing the Earth, so they keep falling towards it but never hitting it. This continual state of free-fall creates the feeling of weightlessness and is why astronauts and objects inside the shuttle appear to float.

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Answer:

Radius of curvature of the path is 1063 meters

Explanation:

It is given that,

Force acting on the pilot is about seven times of his weight. Speed with which pilot moves, v = 250 m/s.

As per Newton's second law of motion, the net force acting on the pilot at the bottom is given by :

[tex]N-mg=\dfrac{mv^2}{r}[/tex]

Where

N is the normal force

r is the radius of curvature

According to given condition,

[tex]7mg-mg=\dfrac{mv^2}{r}[/tex]    

[tex]6mg=\dfrac{mv^2}{r}[/tex]    

[tex]r=\dfrac{mv^2}{6mg}[/tex]  

[tex]r=\dfrac{mv^2}{6mg}[/tex]      

[tex]r=\dfrac{v^2}{6g}[/tex]      

[tex]r=\dfrac{(250)^2}{6\times 9.8}[/tex]  

r = 1062.92 meters

or

r = 1063 meters

So, the radius of curvature of the path is 1063 meters. Hence, this is the required solution.

When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.

Answer: Option C

Explanation:

We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.

But for some materials, like when we rubbed a balloon with human hair, we observed clouding of excess static charge on the balloon surface. This indicates that hair can easily give up electrons to balloon leading to clouding of excess static charge on it.

The electric potential at a point within a uniformly charged solid sphere can be calculated using the given formula. By substituting the given values of charge density, radius of the sphere, and distance from the center of the sphere, the electric potential can be computed.

The potential Vr within a uniformly charged solid sphere of charge density ρ, radius R, at a distance r from the center is given by the formula:

Vr = ρr²/6ε₀ + (ρR²/2ε₀) [ 1 - 3r²/R² ] for r < R

Where ρ is the charge density (4.40*10⁻⁷ C/m³), r is the distance from the center of the sphere (0.160 m), R is the radius of the sphere (0.370 m) and ε₀ is the permittivity of free space (8.854 * 10⁻¹² C²/N.m²).

Substituting the given values into the formula, we can calculate the electric potential at a point located at r = 0.160 m from the center of the sphere.

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The potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

The potential  at a point located at a distance \(r\) from the center of a uniformly charged solid sphere can be calculated using the formula:

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

 where \(k\) is the Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(Q_{enc}[/tex]\) is the charge enclosed within the radius \(r\), and \(r\) is the distance from the center of the sphere to the point where the potential is being calculated.

 For a uniformly charged sphere, the charge enclosed within a radius \(r\) is given by:

[tex]\[ Q_{enc} = \frac{4}{3}\pi r^3 \rho \][/tex]

where [tex]\(\rho\)[/tex] is the charge density of the sphere.

 Given that [tex]\(\rho = 4.40 \times 10^{-7} \, \text{C/m}^3\)[/tex] and [tex]\(r = 0.160 \, \text{m}\),[/tex] we can calculate \(Q_{enc}\) as follows:

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.160 \, \text{m})^3 (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.004096 \, \text{m}^3) (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (1.8192 \times 10^{-9} \, \text{C}) \][/tex]

[tex]\[ Q_{enc} = 2.42528 \times 10^{-9} \, \text{C} \][/tex]

 Now, we can calculate the potential [tex]\(V_r\):[/tex]

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

[tex]\[ V_r = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) (2.42528 \times 10^{-9} \, \text{C})}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = \frac{2.17836 \times 10^{-8} \, \text{Nm}^2/\text{C}}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = 1.36148 \times 10^{-7} \, \text{V} \][/tex]

 Therefore, the potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

Answer:

(a)[tex]h=5.95m[/tex]

(b) h is the same

Explanation:

According to the law of conservation of energy:

[tex]E_i=E_f\\U_i+K_i=U_f+K_f[/tex]

The skier starts from rest, so [tex]K_i=0[/tex] and we choose the zero point of potential energy in the end of the ramp, so [tex]U_f=0[/tex]. We calculate the final speed, that is, the speed when the skier leaves the ramp:

[tex]mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}[/tex]

Finally, we calculate the maximum height h above the end of the ramp:

[tex]v_f^2=v_i^2-2gh\\[/tex]

The initial vertical speed is given by:

[tex]v_i=vsin\theta[/tex]

and the final speed is zero, solving for h:

[tex]h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m[/tex]

(b) We can observe that the height reached does not depend on the mass of the skier

Answer:

Elevation =31.85[m]

Explanation:

We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.

We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.

The point where the ball is thrown shall be taken as a reference point of potential energy.

[tex]E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J][/tex]

m = mass of the ball = 300 [gr] = 0.3 [kg]

v = initial velocity = 25 [m/s]

[tex]E_{k}=\frac{1}{2}  * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J][/tex]

[tex]93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m][/tex]

Answer:

A surface that is full of craters, very similar to that one seen in mercury.

Explanation:

Mercury is the smaller planet in the solar system whit a very thin atmosphere. If Mars were the same size of mercury it will certainly have a surface full of craters.

When meteoroid (fragments of rock) reach the atmosphere of a planet, they get incinerate as a consequence of the friction between the object and the atmosphere. However, if the meteoroid has the necessary size it can reach the ground (at this point is known as meteorite).

In the case propose, it is most likely that bigger meteorite reach the surface of the planet, which leads to the result of a surface full of craters since the size of mars is not enough to maintain a very dense atmosphere due to the weak gravitational field.

Answer:

(a) The greater the frequency of the incident light is, the greater is the stopping potential.

Explanation:

The stopping potential is crucial to terminate the braking of the photoemitted electrons, stopping the current completely.

Since the kinetic energy of electrons depends on the light's incidence frequency (rather than the intensity), therefore, the stopping potential is proportional to the light's incidence frequency.

The cutoff frequency, in turn, is a limiting frequency below which no  photoelectric effect occurs. The cutoff frequency depends on the material from which it is made the emitting surface (and its work function).

Answer:

Part A

Mass = 50g

Vmax = 3.2m/s

Amplitude= 6cm

Position x from the equilibrium= 5.1cm

Part B

Kinetic energy = 0.185J

Part C

Potential energy = 0.185J

Explanation:

Kinetic energy = 1/2mv×2

Vmax = wa

w = angular velocity= 53.33rad/s

Kinetic energy = 1/2mv^2×r^2 = 0.185J

Part c

Total energy = 1/2m×Vmax^2= 0.256J

1/2KA^2= 0.256J

K= 142.22N/m (force constant)

Potential energy = 1/2kx^2

=1/2×142.22×0.051^2

= 0.185J

To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.

In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.

To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.

Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.

Substituting the known values into the equations, we can calculate the kinetic energy of the toy.

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Answer:

-5.985*10^7Ns=momentum

ds=31.578m

Explanation:

The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?

from newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied

f=kdv/dt

k=1

ft=dv/dt

dp=dv/dt

f=negative because its moving in the opposite direction

2.85× 10^6*21=

-5.985*10^7Ns=momentum

-dp=mv2-mv1

-dp+mv1=mv2

-5.985*10^7+2.0× 10^7 kg*3.0 m/s=2.0× 10^7kgV2

v2=150000/(2.0× 10^7)

v2=0.0075m/s to the right

ds=1/2(v2+v1)dt

ds=0.5*(3.0075)*21

ds=31.578m

the barge will travel 31.578m before it is brought to rest.

Answer:

∆p = –5.985×10⁷ kg×(m/s)

∆x = 31.57875 m

The boat's change in momentum is –5.985×10⁷ kg×(m/s), and it will travel 31.57875 meters before it stops.

Explanation:

First off, let's take a quick look at the formula for momentum: momentum equals the force times the time the force is applied for. This means:

p = FΔt

Why is momentum represented by the letter p? That's actually a good question, to be honest—I guess they were running out of letters because the letter m is already used for "mass". But I digress.

Let's go back to the equation. We have both those values—the force applied and the time it is applied for—and so we can substitute those into the equation. But be careful: here, the force is applied in the opposite direction as the actual motion. This means the force must be negative. We get:

p = (–2.85×10⁶ N)(21 s)

p = –5.985×10⁷ kg·(m/s)

That's our value for initial momentum, not the final momentum. Why? Momentum and impulse are equivalent and their units are the same. If we calculated this as the impulse (J = mΔv), we'd be using the initial velocity, not the final velocity (the velocity after the force is applied).

But this problem asked us to solve for two things: the change in momentum (Δp) and the stopping distance (Δx). We can't find either value if we don't know the velocity of the block after the force is applied. This means we need to solve for it.

This is where knowing impulse and momentum are equal comes in handy: if the two values are equal, their formulas should be equal, too. It's like solving a system of equations with the substitution method. With the impulse and momentum formulas, we make them equivalent and get:

mΔv = FΔt

The change in something is like the difference in subtraction: it's the final value minus the initial value. We can then rewrite the equation as:

m(v₂ – v₁) = FΔt

Here, I used (v₂) for the final (second) velocity and (v₁) for the initial (first) velocity. Now, since we need to find the final velocity, (v₂), we need to isolate it by solving the equation for this value.

m(v₂ – v₁) = FΔt

[m(v₂ – v₁)] ÷ m = (FΔt) ÷ m

v₂ – v₁ = (FΔt) ÷ m

v₂ – v₁ + v₁ = [(FΔt) ÷ m] + v₁

v₂ = [(FΔt) ÷ m] + v₁

That's our formula to find the final velocity. The problem already gave us all the values we need to solve this equation: the force applied, the time the force is applied for, the mass of the object being stopped, and the object's initial velocity. When we substitute those into the equation, we get:

v₂ = [(FΔt) ÷ m] + v₁

v₂ = [(–2.85×10⁶ N)(21 s) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = [(–5.985×10⁷ kg·(m/s)) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = 3.0 (m/s) – 2.9925 (m/s)

v₂ = 0.0075 (m/s)

That gives us the final velocity that we need to find (Δp) and (Δx). Now, using the equations for each value, we can substitute in and finish this up!

Δp = m(Δv) = m(v₂ – v₁)

Δp = (2.0×10⁷ kg)[0.0075 (m/s) – 3.0 (m/s)]

Δp = (2.0×10⁷ kg)[–2.9925 (m/s)]

Δp = –5.985×10⁷ kg·(m/s)

Δx = ¹/₂(v₂ + v₁)(Δt)

Δx = ¹/₂[0.0075 (m/s) + 3.0 (m/s)](21 s)

Δx = ¹/₂[3.0075 (m/s)](21 s)

Δx = [3.0075 (m/s)](10.5 s)

Δx = 3.157875 m

And there we go! Problem solved! I hope this helps you! Have a great day!

Final answer:

The question is related to determining the critical value of x, or xcritical, that ensures a bar with a heavy block remains stable by balancing torques around the pivot point.

Explanation:

The question asks about stability conditions in a physical system involving a block on a bar and requires the application of concepts such as the center of mass, force equilibrium, and tension. When the mass of the block is too large and positioned too close to the left end of the horizontal bar, the system may become unstable, potentially causing the bar to tilt. To determine the smallest value of x, denoted as xcritical, we must set up equations considering the torques about the pivot point, ensuring that the sum of torques must equal zero for stability. This involves calculating the torques due to the weight of the bar, the weight of the block, and any other forces acting on the system. The position of the center of mass of the bar is crucial as it influences the bar's propensity to rotate and the distribution of weight. Hence, xcritical is that specific distance at which the torques balance each other out, and the system remains in a stable, horizontal state.

Answer:

[tex]R= \frac{1job}{1.429 hours}[/tex]

So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.

Explanation:

If we want to express the situation in math terms and find the number of hours that takes to complete 1 job with the 3 at the same time, we can do this.

For this case we have the following rates:

[tex]R_A =\frac{1job}{6hours}[/tex]

[tex]R_B =\frac{1 job}{5 hours}[/tex]

[tex]R_C=\frac{1job}{3 hours}[/tex]

And we know that working together the rate would be the addition of the rates like this:

[tex]R=R_A +R_B +R_C = \frac{1job}{6hours}+\frac{1job}{5hours}+\frac{1job}{3hours} =\frac{7 jobs}{10hours}[/tex]

And if we divide the numerator and denominator by 7 we got:

[tex]R= \frac{1job}{1.429 hours}[/tex]

So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.

Answer:

The Potential difference across the 8 Ω resistor = 9.6 V

Explanation:

In a series combination of resistor, The total resistance

Rt = R₁ + R₂............... Equation 1

Where Rt = total resistance, R₁ = Resistance of the first resistor, R₂ = resistance of the second resistor.

Given: R₁ = 12Ω, R₂ = 8Ω.

Substituting these vales into equation 1

Rt = 12 + 8

Rt = 20 Ω.

Since both resistors are connected in series, The same current flows through both resistors.

from ohms law,

V = IR..................... Equation 3

I = V/R....................... Equation 2

where I = current flowing through both resistors, V= Emf, R = total resistance of the circuit

Also given : V = 24 V, R = 20 Ω,

Substituting these values into equation 3,

I = 24/20

I = 1.2 A.

Note: Since both resistors are connected in series, The same current flows through both resistors.

The Potential difference across the 8 Ω resistor

V = 1.2×8

V = 9.6 V

The Potential difference across the 8 Ω resistor = 9.6 V