How does changing the frequency affect the wave speed

The speed of the chain at this instant is 8.47 m/s. The change in energy of the chain is 204.8J. The increase in thermal energy of the chain is 25.6J.

Acceleration(a) of Centre of mass = F/m = 64/5 = 12.8 m/s2

a) v2 = u2 + 2as,....(1)

v is final velocity of centre of mass,

u is the initial velocity of centre of mass,

s is the displacement of centre of mass

s = d - L/2 = 3.2 - 0.4 = 2.8m

using equation 1,

V = \(=\sqrt{2as} =\sqrt{2*12.8*2.8}=8.47 m/s.\)

Therefore, the speed of the chain at this instant is 8.47 m/s.

2. Change in energy = Total Work Done = Force*Displacement of point on which force is applied = 64*3.2 = 204.8J

Therefore,  the change in energy of the chain is 204.8J

3. Change in Thermal Energy = Change in Energy - Change in Kinetic Energy = 204.8 - 0.5*m*v2

Change in Thermal Energy = 204.8 - 0.5*5*8.47*8.47 = 25.6J

Therefore, the increase in thermal energy of the chain is 25.6J.

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(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that

0² - v² = 2 a d → a = - v² / (2d)

Then the car comes to rest over a distance of

d = - v² / (2a)

Doubling the starting speed gives

- (2v)² / (2a) = - 4v² / (2a) = 4d

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration a such that

0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass m such that

60 J = m g h

where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives

m = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

Answer:

Explanation:

Compete question;

A roller coaster has a loop in which the centripetal acceleration equals 9.8m/s² . If the tangential speed of the roller coaster cars is 15.7 m/s, find its radius.

centripetal acceleration a = v²/r

v is the tangential speed

r is the radius

Given

v = 15.7m/s

a = 9.8m/s²

9.8 =  15.7²/r

9.8r = 15.7²

r = 15.7²/9.8

r = 246.49/9.8

r = 25.152m

Hence the radius of the coaster is 25.152m

The statements true in regards to heat is 3. Heat is a form of energy, can be reflected by a mirror, and cannot pass through a vacuum.

Heat is a form of energy. Heat is the transfer of thermal energy from one object to another. Thermal energy is the energy of motion of the particles in an object. Heat can be reflected by a mirror. Heat is a form of electromagnetic radiation, and electromagnetic radiation can be reflected by mirrors.

Heat cannot pass through a vacuum. Heat is a form of electromagnetic radiation, and electromagnetic radiation cannot pass through a vacuum. Heat is not an electromagnetic radiation. It is a form of energy that is transferred from one object to another because of a difference in temperature.

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Complete question:

Which of the following statements are true regarding heat?

(a) Heat is a form of energy

(b) Heat can be reflected by mirror

(c) Heat is an electromagnetic radiation

(d) Heat can pass through vacuum

Select the correct answer from the codes given below :

1. 1, 2 and 3

2. 2, 3 and 4

3. 1, 2 and 4

4. 1, 3 and 4

The substance will be reduced to 1/16 of A present, that is 1 lb, in 95 hours.

Let the initial amount of A present be X lb. The rate of decomposition of A is proportional to the amount of A present. Therefore, the rate of decomposition = k * X where k is the proportionality constant. We know that 1 lb of A will reduce to 4 lb in 38 hours. So, the rate of decomposition = X/38.

Also, the rate of decomposition = k * X. Comparing both the equations, k = 1/38. Therefore, the rate of decomposition = X/38A substance will reduce to 1/16 of A present i.e., X/16. Using the equation for the rate of decomposition, we get, X/16 = (1/38)*X*(t). Simplifying, we get t = 95 hrs. Hence, the substance will be reduced to 1/16 of A present, that is 1 lb, in 95 hours.

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Answer:

P V = N R T      Ideal gas equation

1 mole CO2 = 12 + 32 = 44 g

5 g = 5 / 44 = .114 moles

One definition of R is R = .08207 L-Atm / mole-deg K

P = .114 * .08207 * 298 / 10 = .279 Atmosphere

One needs to use care in using R in terms of other values given

3.5 s is the  time elapses from when the rock is thrown until it hits the street when thrown vertically upward.

Every area of our lives is impacted by the concept of time. It determines the duration of an event and the tempo of our daily activities. Time begins to move forward as soon as we open our eyes in the morning, reminding us of the obligations we have to fulfil and the dates we have to meet. We use time to organise our lives, prepare our calendars, communicate with others, and plan events. Time is an elusive and philosophical term that defies simple explanations despite the many ways it forms and affects our life. It passes in the blink of an eye while also appearing to go on forever. It is both finite and endless.

\(mgh = \frac{1}{2}mv^2 + mgh_{\text{roof}}\)

v =\(\sqrt{2gh}\)

  = \(\sqrt{2(9.81\ \text{m/s}^2)(28\ \text{m})} \approx \boxed{18.8\ \text{m/s}}\)

h =\(\frac{1}{2}gt^2\)

t =\(\sqrt{\frac{2h}{g}}\)

=\(\sqrt{\frac{2(28\ \text{m})}{9.81\ \text{m/s}^2}} \approx \boxed{3.5\ \text{s}}\)

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The ratio of the ideal projectile distance on the moon to the same projectile distance on earth, moon/back, is approximately (the acceleration of gravity on the moon is 1.62 m/s², about one-sixth of the acceleration due to gravity on earth. for a given initial velocity v0  , and the specified launch angle θ0) is ⅙ (option B)

The initial velocity, launch angle, and acceleration due to gravity all affect a projectile's range. The formula for an ideal projectile's range is as follows:

R = v0²sin (2θ0) / g

Where,

v0 = the initial velocity (m/s)

θ0 = the launch angle

g = the acceleration due to gravity

The acceleration caused by gravity is 1.62 m/s² for the moon and 9.8 m/s² for the earth. Ranges on the moon and ranges on earth can be expressed as follows:

R moon / R earth = (1.62 sin (2θ0)) / g

sin (2θ0) = 1.62 / 9.8

Approximately ⅙

Consequently, an ideal projectile's range on the moon is around a sixth of its range on Earth.

The question is incomplete, it should be:

The acceleration due to gravity on the moon is 1.62 m/s², approximately a sixth of the value on earth. for a given initial velocity v0, and a given launch angle θ0, the ratio of the range of an ideal projectile on the moon to the range of the same projectile on earth, rmoon/rearth, will be approximately?

A. ⅓

B. ⅙

C. 1

D. 1/12

E. ⅕

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The velocity-time graph for the entire motion of the lift can be divided into three sections as given below:

a) SECTION A: The lift moves with an initial velocity of 0 m/s, so the graph starts from (0,0) point. It moves downwards for a distance of d metres with a constant acceleration of g/2 m s−2.

SECTION B: After moving the first distance d metres, the lift moves with the velocity attained at the end of the previous motion. This means that the velocity of the lift is constant for this section, so this section of the graph is a straight line parallel to the x-axis.

SECTION C: In this section, the lift moves the remaining distance of 2d metres with a constant deceleration. So the graph will be a straight line making an angle of θ with the time axis.

(b) Let the total time taken by the lift to move down from A to B be t sec.

Therefore, time taken by the lift in section A = time taken by the lift in section B= t/2 sec

Let the speed attained by the lift at the end of section A be V m/s. So the velocity of the lift at the end of section B is also V m/s.

In section B, the distance covered by the lift = d metres.Time taken by the lift in section

C = t - (t/2 + t/2) = t/2 sec

Distance covered by the lift in section C = 4d - 2d = 2d metres

As the lift comes to rest at the end of section C, we have,

Final velocity of the lift, v = 0 m/s

Initial velocity of the lift, u = V m/s

Acceleration of the lift,

a = - g/2 m s−2

Distance covered by the lift, s = 2d m

Using the second equation of motion, we have

v = u + at0 = V + (-g/2)t/2V = g/4 t ----(1)

Using the third equation of motion, we have

2d = Vt/2 - (g/4)(t/2)²2d

= Vt/2 - (gt²)/32d = Vt/2 - (gt²)/8

Substitute V from equation (1)

2d = (g/4)t(t/2) - (gt²)/8

2d = (gt³)/16t³ = 32dt = 2∛(4d)

Therefore, the total time taken by the lift to move down from A to B is 2∛(4d) sec.

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The polarization axes of glasses for 3-d viewing are in vertical and horizontal.

vertical and horizontal. The 3-d glasses have two lenses. Each lens is polarized in a different way. The lenses are aligned to the image on the screen and each lens allows only one image to enter the eye. This creates an illusion of depth or 3-Dimensional. The light waves that make up an image are polarized. In 3-D glasses, each lens has a different polarization. The lenses allow only the image intended for that eye to be seen. Hence, the brain can integrate the two images to form a single 3-D image.

3-D glasses create an illusion of depth by creating two different images. Each lens of the glasses is polarized in a different way, either vertically or horizontally. One of the lenses will allow the vertical polarization to be seen by one eye and the other lens will allow only horizontal polarization to be seen by the other eye. The two different images are sent to the brain, and it creates an illusion of depth. The polarization of the lenses blocks out the light from the opposite polarization. The images on the screen are transmitted through two different projectors, which are polarized at different angles. The images are projected onto the screen, and the polarization of the lenses filters the images.

3-D glasses work by polarizing the lenses in different ways to allow each eye to see a different image, which creates an illusion of depth. The polarization axes of glasses for 3-d viewing are vertical and horizontal.

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Hi there!

We can begin solving for the work necessary using the following:

\(W = \Delta U = mg\Delta h\)

W = Work (J)

Δh = change in height (10 m)
m = mass (75 kg)
g = acceleration due to gravity (9.8 m/s²)

Plug in the givens:

\(W = 75 \cdot 9.8 \cdot 10 = 7350 J\)

Now, we can solve for power with this equation:
\(\large\boxed{P = \frac{W}{t}}\)

P = Power (Watts)

W = Work (J)
t = time (s)

Plug in the values:

\(P = \frac{7350}{25} = \boxed{294 W}\)

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

                           \(w=\frac{2*wavelength*d}{a}\)

where, d is the distance between slit and the screen, and a is the slit width.

                           \(d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m\)

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

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The total distance you would have covered is equivalent to going around the earth 22 times.

The total distance traveled at a given speed and time of motion is calculated as follows;

Distance = speed x time

\(Distance = \frac{25 \ miles}{day} \times \frac{365 \ days}{1 \ year} \times 60 \ years\\\\Distance = 547, 500 \ miles\)

The distance round the earth or circumference of the earth of the earth has been given as 25,000 miles

\(n = \frac{547,500}{25,000} \\\\ n = 21.9 \ times \ \approx 22 \ times\)

Thus, the total distance you would have covered is equivalent to going around the earth 22 times.

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The internal resistance is 3.92 Ω.

An ideal emf of 19 volts, voltage across the terminals is 13.2 volts, current is 1.47 amps.

Resistance and voltage:

In terms of power, voltage is defined as the work done per unit charge across two points in an electric circuit. It is measured in volts (V).

Resistance is defined as the opposition to the flow of electric current through a conductor. It is measured in ohms (Ω).

Internal resistance:

Let R be the internal resistance.

The voltage, V = 19 V - IR

The voltage across the terminals, V' = 13.2 V, and the current, I = 1.47 A.

The equation for calculating internal resistance is given asR = (V - V') / I

Substitute the given values.

R = (19 - 13.2) / 1.47R = 3.92 Ω

Therefore, the internal resistance is 3.92 Ω.

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a simple lifting machine consisting of a rope which unwinds from a wheel on to a cylindrical drum or shaft joined to the wheel to provide mechanical advantage. reeeeeeeeeeeeeeeeeeeeeeeeee

Answer:

Newton was a mathematician, physicist and more. He co-invented calculus. But I have no idea about the atom part.

The charge carried by the small sphere can be determined using Newton's second law and the given acceleration.

According to Newton's second law, the force acting on an object is equal to its mass multiplied by its acceleration. In this case, the electric field causes the sphere to acquire a horizontal acceleration of 13.0 m/s^2.

The force acting on the sphere can be calculated using the formula F = m * a, where F is the force, m is the mass, and a is the acceleration. Plugging in the values, we have F = (0.441 kg) * (13.0 m/s^2) = 5.733 N.

The electrostatic force experienced by the sphere is given by the equation F = q * E, where q is the charge and E is the electric field strength. Rearranging the equation, we can solve for the charge: q = F / E. Substituting the values, we get q = (5.733 N) / (5.00 N/C) = 1.147 C.

Therefore, the charge carried by the small sphere is 1.147 Coulombs.

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Answer:

23.11

Explanation:

accellus

The focal length of the lens is approximately 115.3 cm. The explaination is  explained below.

To find the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where:

f is the focal length of the lens

v is the distance of the image from the lens

u is the distance of the object from the lens

Given:

v = 38.5 cm

u = 57.8 cm

Let's substitute these values into the formula:

1/f = 1/38.5 - 1/57.8

Now, let's calculate the right side of the equation:

1/f = (57.8 - 38.5) / (38.5 * 57.8)

= 19.3 / (2223.3)

≈ 0.008677

Now, we can find the focal length (f) by taking the reciprocal of both sides:

f = 1 / (0.008677)

≈ 115.3 cm

Therefore, the focal length of the lens is approximately 115.3 cm.

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The system being referred to here is likely the swim bladder of a fish. This organ helps a fish maintain buoyancy in the water column by adjusting the amount of gas it contains. By changing the volume of the gas-filled swim bladder, a fish can move up or down in the water column as needed.

Similarly, a fish can adjust its position in the water column by swimming up or down using its fins. This is necessary for survival as different species of fish have different feeding and habitat requirements. When fishermen bring a fish from great depths to the surface quickly, the sudden change in pressure can cause the swim bladder to expand rapidly and even rupture. This is known as barotrauma and can be fatal for the fish. To prevent this, fishermen can use techniques such as venting or releasing the fish at a slower rate to help it acclimate to the changing pressure.

If a fish is prohibited from keeping or the fisherman simply wants to release it, they can use a technique called a descending device. This is a tool that helps the fish return to its habitat on the bottom safely and quickly. The device typically consists of a weight and a release mechanism that allows the weight to drop the fish back down to the desired depth. This helps the fish avoid the stress of being caught and brought to the surface and increases its chances of survival.

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Answer:

b

Explanation:

False.

The continuity equation relates the mass flow rate (ρAv) at one point in a flow to the mass flow rate at another point in the flow. It is given by:

ρ₁Av₁ = ρ₂Av₂

where ρ is the density, A is the cross-sectional area, and v is the velocity.

In the case of steady flow, the mass flow rate at any point in the flow is constant, which means that ρAv is constant.

Therefore, the continuity equation reduces to:

ρAv = constant

This equation can be rearranged to give:

A₁v₁ = A₂v₂

which is the equation of continuity for incompressible fluids.

The assertion of the divergence-free property of a vector field, where the vector field's divergence is zero, is made by the equation laplacian operator * v = 0, which has nothing to do with the continuity equation.

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Yes, it is possible for an object to be moving in one direction while the net force acting on it is in another direction. This is because the object's motion is determined by the combination of all the forces acting on it, not just the net force.

For example, consider a car driving on a circular track at a constant speed. The net force acting on the car is directed towards the center of the circle, which is perpendicular to the car's direction of motion. However, the car continues to move forward due to the inertia of its motion. Another example is a rocket moving in space. The rocket's engines apply a force in one direction, but the rocket can continue to move in another direction if it has sufficient velocity and is not acted upon by any external forces. In both cases, the object's motion is not solely determined by the net force acting on it, but also by its initial velocity and the absence of any other external forces.

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The equation of the force between the two masses separated by distance r, is determined as Gmm/r².

The equation of the force between the two masses is determined from Newton's law of universal gravitation as shown below.

f = Gmm/r²

where;

Thus, the equation of the force between the two masses separated by distance r, is determined as Gmm/r².

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Answer:

Approximately \(722\; \rm m\cdot s^{-1}\).

Explanation:

The average speed of a vehicle is calculated as:

\(\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}\).

In this question, the total distance is \(2 \times 560\; \rm km = 1120\; \rm km\).

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

\(560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m\).

\(1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m\).

Time required for the first part of this trip:

\(\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s\).

Time required for the second part of this trip:

\(\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s\).

The time required for the entire trip would be approximately \(930 + 620 = 1550\; \rm s\).

Calculate the average speed of this plane:

\(\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}\).

Answer:

what are the options?

Explanation:

The labelled table data hat the experimenter can record observations for the sand and water temperatures at various points is given below.

Here is a labeled data table for recording sand and water temperatures at various points:

Point            Sand Temperature (°C)                 Water Temperature (°C)

1  

2  

3  

4  

5  

The table is 5 columns wide and 3 rows long, with the first column labeled "Point" to indicate the location being observed, and the second and third columns labeled "Sand Temperature (°C)" and "Water Temperature (°C)" respectively to indicate the type of temperature being measured.

The cells under the "Sand Temperature (°C)" and "Water Temperature (°C)" columns are left blank to allow the experimenter to record the corresponding temperature readings for each point.

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Full Question:

Although part of your question is missing, you might be referring to this full question:

Label a data table so that the experimenter can record observations for the sand and water temperatures at various points.

the column table 5 length x 3 width

We are given the following point:

Part A. To determine the quadrant of the point we will plot the point in a coordinate plane:

Therefore, point A is in quadrant I.

Part B. we are given the following transformation rule:

Applying the rule:

Answer:

P = 0.2 W

Explanation:

Given that,

Force, F = 4 N

Displacement, d = 1 m (say)

Time, t = 20 s

We need to find the power output. Power is the rate of doing work. So,

P = W/t

\(P=\dfrac{Fd}{t}\\\\P=\dfrac{4\times 1}{20}\\\\P=0.2\ W\)

So, the required power is 0.2 W.

Answer:

infrared will do but I'm not sure this question is hard if I may say and we cannot take green because it is the colour that the concept must have