The geologic time scale originally ordered earth’s rocks by

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:

(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:

The time taken by the projectile to fall to the ground is 1 second.

Explanation:

It is given that,

Acceleration due to gravity on the planet Zorb, [tex]a=10\ m/s^2[/tex]

Angle of projectile is 30 degrees

Initial velocity of the projectile, u = 10 m/s

Let t is the time taken for the projectile to fall to the ground. Using second equation of motion to find it

[tex]h=u\ sin\theta t+\dfrac{1}{2}at^2[/tex]

When it fall to the ground, h = 0 and a = -g

[tex]10\ sin(30)t-\dfrac{1}{2}\times 10t^2=0[/tex]

On solving the above quadratic equation, we get the value of t as,

t = 1 second

So, the time taken for the projectile to fall to the ground is 1 second. Hence, this is the required solution.

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

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 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:

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:

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:

Transform active margins are associated with which type of boundary?

Transform boundary

Explanation:

The transform boundary is a boundary where one plates(crust) slides past another plate horizontally. This kind of plate movement have been detected to exist between the interaction of the North pacific plates(continental plate) and the pacific plates(oceanic plates) .

At the transform margin the crust are usually broken. But overall crust are neither created nor destroyed . The transform margin region are active as it is marked by shallow-focus earthquakes .

Along the fractured zone where this transform movement occurs is known to create extensive transform faults .Notable transform fault that exist in this kind of boundary(transform) is the San Andrea fault and Alpine Fault.  

The motion of this plates can occur on a single fault or on a group of faults.

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:

Explanation:

During any collision whether elastic or inelastic, force of action and reaction are generated at the point of touch for a brief period during which they remain in touch with each other. These action and reaction forces are equal and opposite. Impulse is defined by force multiplied by time duration . Hence we can say that during any collision , the impulse generated are equal and opposite .

In inelastic collision , colliding objects coalesce with each other . Hence in the given case , collision is inelastic.

In elastic collision , when an object collides with a similar object at rest , there is exchange of velocity . In the given case also , similar objects are colliding and exchange of velocity is taking place.So it is an example of elastic collision.

In an inelastic collision , momentum is conserved. So final total momentum is equal to initial total momentum.

When a cannon fires a cannonball, the cannon will recoil backward because the total momentum is conserved.  

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

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:

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:

Tire blowout

Explanation:

Tire Blowout

A tire experiences a blowout when it rapidly and explosively loses the inflation pressure. It can happen when some sharp object cuts or tears the surface and the air is suddenly released, making things worse.

The driver frequently loses control of the traveling direction and can be involved in serious accidents. It is recommended not to use brakes during or after the blowout because it could bring little help or make things worse.

Focusing on steering and let the vehicle softly brake.

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:

The space in seconds that will be kept = 2.27 seconds

Explanation:

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

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

Conversion: (i)if 1 mph = 0.44704 m/s,

                then, 30 mph = 30×0.44704    

               = 13.41 m/s

              (ii) If 1 foot = 0.3048 m

           then, 100 foot = 30.48 m.

Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m

Substituting these values into equation 2

t = 30.48/13.41

t = 2.27 seconds.

Therefore the space in seconds that will be kept = 2.27 seconds

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.

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:

The charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively.

Explanation:

It is given that,

Radius of the conducting sphere, r = 0.15 m

Potential, V = 200 V

Potential on the surface of sphere is given by :

[tex]V=\dfrac{kq}{r}[/tex]

q is the charge on the sphere

[tex]q=\dfrac{Vr}{k}[/tex]

[tex]q=\dfrac{200\times 0.15}{9\times 10^9}[/tex]

[tex]q=3.34\times 10^{-9}\ C[/tex]

Charge per unit area is called charge density on the surface. it is given by :

[tex]\sigma=\dfrac{q}{A}[/tex]

[tex]\sigma=\dfrac{q}{4\pi r^2}[/tex]

[tex]\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}[/tex]

[tex]\sigma=1.18\times 10^{-8}\ C/m^2[/tex]

So, the charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively. Hence, this is the required solution.

Final answer:

The charge on the conducting sphere with a radius of 0.15 m and a potential of 200V is approximately 3.34 × 10⁻⁶ C, and the surface charge density is approximately 1.18 × 10⁻⁵ C/m².

Explanation:

To find the charge (q) and the surface charge density (σ) on the surface of a conducting sphere with a given potential (V), we can use the relationship between potential, charge, and radius of a sphere, along with the equation that relates charge to surface charge density.

The potential (V) of a conducting sphere is given by V = (k × q)/R, where k is Coulomb's constant (k = 8.99 × 10⁹ N.m²/C²), q is the charge on the sphere, and R is the radius of the sphere.

In this case, V = 200V and R = 0.15m. Using the formula, we can find the charge q on the sphere:

V = (k × q)/R
200V = ((8.99 × 109 N.m²/C²) × q)/0.15m
q = (200V × 0.15m)/(8.99 × 10⁹ N.m²/C²)
q ≈ 3.34 × 10⁻⁶ C

Once we have q, we can use the equation q = σ(4πR2) to find the surface charge density σ:

σ = q / (4πR²)
σ ≈ 3.34 × 10⁻⁶ C / (4π × (0.15m)²)
σ ≈ 1.18 × 10⁻⁵ C/m²

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:

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: 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:

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

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:

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:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

[tex]-h= vsinA\times t-gt^2/2[/tex]

putting values h=15 m, v=0.8

[tex]-15 = 0.8vt - 4.9t^2[/tex]  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

[tex]-15 = 0.8\times40/0.6 - 4.9t^2[/tex]

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = [tex]-H =v sinA t - gt^2/2[/tex]

⇒ [tex]4.9t^2 - 8.9\times0.8t - 100 = 0[/tex]

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

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:

(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:

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