A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attached to the end of the unstretched spring and then released. The fish moves downward until the spring is fully stretched, then starts to move back up as the spring begins to contract.
What is the maximum distance through which the fish falls?

Answer:

the answer is the first one.

Explanation:

one way to find out is if you look at the other equations they are mixed matched. For an example a=d/t should be s= d/t.

Answer:

a

Explanation:

i did it

Answer:

19230 kg/m^3

Explanation:

Divide its mass by its volume to get its density:

500 kg/0.026 m = 19230 kg/m^3

Also, please provide the question next time since other's didn't know what they should solve for in this problem-- I only knew because I had this same problem.

Answer:

19230 kg/m3

Explanation:

I took the test, that is the answer

A fast meat eating dinosaur  must have strong claws for catching prey as well as strong teeth for eating meat.

Dinosaurs are ancient reptiles that occurred in different forms and shapes and are currently considered a part of the evolutionary sequence.

A carnivorous dinosaur must have strong claws for catching prey as well as strong teeth for eating meat.

The image of a carnivorous dinosaur is shown in the image attached to this answer.

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Answer:

.5

Explanation:

First we define the variables

v=20

r=80

Next we construct our formula, since our tires do not slip, our centripetal force is equal to our friction

ma=mg*u

We calculate our centripetal force through finding our acceleration (v^2/r) and assuming our mass is a variable m (we cancel it out later).

mv^2/r = mg*u

All we have left is substitution and simplification

400m/80=mg*u

5=g*u

G is our gravitational acceleration (I will use 10, use 9.8 instead if you prefer)

5=10*u

u=.5

Answer:

Data that we know:

Mass of the car = 950kg.

Gravitational acceleration = -9.8m/s^2.

Acceleration of the car = -1.2 m/s^2.

(the negative signs are because that accelerations point downwards)

Now, the total acceleration of the car is:

-1.2m/s^2 = -9.8m/s^2 + Ac.

Where Ac is the acceleration caused by the cable, that is opposite to the acceleration of the car:

Ac = 9.8m/s^2 - 1.2m/s^2 = 8.6m/s^2.

Now we have the accelerations, remember that by second's newton law we have:

F = m*a

Then the force of the cable is:

Fc = 8.6m/s^2*950kg = 8,170N.

The gravitational force is:

Fg = -9.8m/s^2*950kg = -9,310N

The net force is:

Ft = 8,170N - 9,310N = -1,140N.

Below is the free body-diagram.

Answer:17m/s

Explanation:

In this question we use first law of motion which is

vf=vi+at

By putting values

vf=32+10×-1.5

vf=17m/s

Answer:

yeah its 17m/s

Explanation:

Answer:

896000 Joule.

Explanation:

Solution,

Given,

Mass (m) = 1,120 kg.

Velocity (v) = 40 m/s.

Now,

Kinetic Energy = 1/2mv²

= 1/2 × 1,120 × 40²

= 896000 Joule.

Answer:

B-A-C

Explanation:

The idea is to evaluate the slopes of the three graphs. Notice that the three graphs are a representation of position (x) as a function of time (t).

Then, the slope of those lines are giving information of the change in position over the change in time (rise over the run). For the graph with positive slope the velocity is positive, for those with negative slope (line going down) the velocity is negative. But we are asked to compare speeds, which are the magnitude  of each velocity (slope) so it is important to determine the graph with the largest slope (graph C), and that with the smallest (graph B)

Then the order is: B-A-C (third option answer option)

I cannot provide the information for the student guide as I do not have access to the specific lab or the guide. So, I cannot confirm whether I have read the student guide or not. Please provide me with the necessary information or a specific lab for me to provide accurate and helpful information. Instead of this I can give a brief about the student guide for a physics lab.

A student guide for a physics lab is a document that provides students with information about the expectations, procedures, and safety guidelines for the lab. The guide usually includes the following information:

1. Introduction: A brief overview of the lab, including its purpose and objectives.

2. Materials and equipment: A list of the materials and equipment that students will use during the lab, as well as instructions on how to use them.

3. Procedure: A step-by-step guide on how to perform the experiment, including any calculations or data analysis that may be required.

4. Safety guidelines: A list of safety guidelines that students should follow during the lab, including proper handling of materials, use of protective equipment, and emergency procedures.

5. Pre-lab assignments: Assignments that students need to complete before the lab, such as reading the lab manual or reviewing relevant concepts.

6. Post-lab assignments: Assignments that students need to complete after the lab, such as analyzing data, writing lab reports, or answering questions.

Therefore, A student guide is an essential tool that helps students prepare for the lab, understand the concepts and procedures involved, and stay safe while conducting the experiment. It is important for students to read the entire guide carefully to ensure they have a clear understanding of the lab expectations and procedures.

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Answer:

acceleration, rate at which velocity changes with time, in terms of both speed and direction. A point or an object moving in a straight line is accelerated if it speeds up or slows down. Motion on a circle is accelerated even if the speed is constant, because the direction is continually changing. For all other kinds of motion, both effects contribute to the acceleration.

Because acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is also a vector quantity. Acceleration is defined as the change in the velocity vector in a time interval, divided by the time interval. Instantaneous acceleration (at a precise moment and location) is given by the limit of the ratio of the change in velocity during a given time interval to the time interval as the time interval goes to zero (see analysis: Instantaneous rates of change). For example, if velocity is expressed in metres per second, acceleration will be expressed in metres per second per second.

Explanation:

Answer:

[tex]v=697.2km/h[/tex]

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

[tex]V=\frac{d}{t}[/tex]

The distance is clearly 1743 km and the time is:

[tex]t=2h+30min*\frac{1h}{60min} =2.5h[/tex]

Thus, the velocity turns out:

[tex]v=\frac{1743km}{2.5h}\\ \\v=697.2km/h[/tex]

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

Answer:

Acceleration = 8.27 cm/s²

Explanation:

We are given;

initial velocity; v_i = 10.5 cm/s

Initial position; x_i = 2.72 cm

Time; t = 2.30 s

final position; x_f = 5.00 cm

To find the acceleration, we will make use of the formula;

x_f - x_i = (v_i * t) - (½at²)

Plugging in the relevant values, we have;

5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)

2.28 = 24.15 - 2.645a

24.15 - 2.28 = 2.645a

2.645a = 21.87

a = 21.87/2.645

a = 8.27 cm/s²

The tension in the wire is caused by the elastic collision between the balls and the magnitude of this tension is determined as 18.81 N.

The speed of the balls after the collision is determined by applying the principle of conservation of linear momentum as shown below.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

9.2(0) + 2.8(3.5) = 9.2v₁ + 2.8(v₂)

9.8 = 9.2v₁ + 2.8v₂   -----(1)

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3.5 + v₂

v₂ = v₁ - 3.5  ---(2)

Substitute (2) into (1)

9.8 = 9.2v₁ + 2.8(v₁ - 3.5)

9.8 = 9.2v₁ + 2.8v₁ - 9.8

19.6 = 12v₁

v₁ = 19.6/12

v₁ = 1.63 m/s

v = √T/μ

v² = T/μ

T = v²μ

where;

μ is mass per unit length of the wire = 9.2/1.3 = 7.08 kg/m

T = (1.63²) x 7.08

T = 18.81 N

Thus, the tension in the wire just after collision is 18.81 N.

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Answer:

the c is in the cycle lab where they were created and then it cycles then it reoves it to the right.

Explanation:

The displacement of the airplane from the take off point after travelling the given distance is 15 km.

The displacement of the airplane is the shortest distance between the initial position of the airplane and the final position of the airplane. The displacement of the of the airplane is calculated as follows;

R² = Rx² + Ry²

R² = 10.4² + 10.8²

R² = 224.8

R = √224.8

R = 15 km

Thus, the displacement of the airplane from the take off point is 15 km.

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Answer:

the correct answer is 125 joule

Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

a = 4.9 m/s²

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

Note: N - m g is the net force producing the centripetal force

Answer:

Pascal

Explanation:

The standard SI unit for pressure measurement is the Pascal (Pa) which is equivalent to one Newton per square meter (N/m2) or the KiloPascal (kPa) where 1 kPa = 1000 Pa.

Answer:

[tex]\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}[/tex]

Explanation:

[tex](inertia.of.the.disk) Id =\frac{M R^{2}}{2}\\(inertia.of.the.solid) Is = m R^{2}\\\\Itotal = Id + Is = \left(\frac{M}{2}+m\right) R^{2}\\\\\\L=\vec{\gamma} \times \vec{p}={mvr} \quad \& \quad L=I \omega\\\\Mrock \timesv \times R=\left(\frac{M}{2}+m\right) R^{2} \times \omega\\\\\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}\\[/tex]

The required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].

Given data:

The mass of disk is, M.

The radius of disk is, R.

The mass of teenager is, m.

The mass of rock is, [tex]m_{rock}[/tex].

The speed of rock is, v.

Here we  need to apply the concept of angular momentum to obtain the value of angular speed of disk. The expression for the angular momentum is given as,

[tex]L = I_{total} \times \omega[/tex] .......................................(1)

Here, [tex]\omega[/tex] is the angular speed and [tex]I_{total}[/tex] is the total moment of inertia of system.

And its value is,

[tex]I_{total} = I_{disk}+I_{solid}\\\\I_{total} = \dfrac{MR^{2}}{2}+mR^{2}[/tex] ..........................................(2)

And the angular momentum is also expressed as,

[tex]L = p \times R\\L =( m_{rock}v) \times R[/tex] ....................................................(3)

Then, using the equation (1), (2) and (3) we have,

[tex]( m_{rock} \times v) \times R = (\dfrac{MR^{2}}{2}+mR^{2}) \times \omega\\\\m_{rock} \times v = (\dfrac{MR}{2}+mR) \times \omega\\\\\\\omega = \dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex]

Thus, we can conclude that the required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].

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Answer:

25 kilometers an hour.

Explanation: Since there is 100 kilometers, you dived it by four hours and you get the kilometers per hour.

Answer:

25 kmph

Explanation:

100 divided by 4 is 25

hope it helps

Answer:

The answer is C:  a 75kg dining table

Explanation:

Took it on the Unit Test: 2020

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N   ⇒   k = (68.6 N) / (0.105 m) ≈ 653 N/m

Answer:

A) v = 31.8 m/s

B) v_f = 18.1 m/s

C) F_avg = 4130.7 N

Explanation:

A) From law of conservation of energy, we know that;

mgh = W_f + ½mv²

v is the speed at which she is going at the bottom of the slope.

Thus, making v the subject, we have;

v = √[2gh - (2W_f)/m)]

We are given;

h = 70 m

m = 58 kg

W_f = 1.04 × 10⁴ J = 10400 J

Thus;

v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]

v = 31.8 m/s

B) Total force will be given by the formula;

F_t = F_k + F_r

Where;

F_k is force of kinetic friction = mg•μ_k

μ_k = 0.25

So, F_k = 58 × 9.8 × 0.25

F_k = 142.1 N

We are given force of air resistance(F_r) as 150 N

Thus;

F_t = 142.1 + 150

F_t = 292.1 N

Final velocity is gotten from the formula;

v_i² - v_f² = 2F_t•L/m

Thus;

v_f = √[v_i² - (2F_t•L/m)]

Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m

Thus;

v_f = √[31.8² - (2 × 292.1 × 68/58)]

v_f = 18.1 m/s

C) the average force exerted on her by the snowdrift as it stops her is given by the formula;

F_avg = m•v_f²/2l

F_avg = 58 × 18.1²/(2 × 2.3)

F_avg = 4130.7 N

Answer:

Answer is attached

The average velocity for the race is 70 m/s if the dragster crosses the finish line with a velocity of 140 m/s.

It is defined as the rate of change in linear velocity with respect to time. It is also known as linear acceleration.

It is given that:

A dragster crosses the finish line with a velocity of 140m/s.

The vehicle maintained a constant acceleration from start to finish.

As we know,

The average velocity can be given as:

V(avrg) = (v1 + v2)/2

v1 = 140 m/s

v2 = 0 m/s

v(avrg) is the average of the velocity.

V(avrg) = (140 + 0)/2

V(avrg) = 140/2

V(avrg) = 70 m/s

Thus, the average velocity for the race is 70 m/s if the dragster crosses the finish line with a velocity of 140 m/s.

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Answer:

The answer is B.

Explanation:

A Clan is usually just one family, but can be multiple families that are related. They were usually very close knit. Sometimes a clan is also a group of people with a common interest.

Answer:

Deposition

Explanation:

When the wind calms down or vegetation stops or slows the breeze, the sediment particles begin to fall. Water is another factor that may erode, move, or deposit sediment. Flowing water is a key erosive agent. Water transports dirt and rock fragments as it moves. Warm, wet air will not travel if wind systems are not present. Water will still evaporate, but it will not move, therefore everywhere else than a major body of water will dry up. Lakes may be fine since evaporating water will flow back into them, and the sea will be fine, but everywhere else will become extremely dry very rapidly. Wind is constantly blowing somewhere on the world at any given time. Winds are usually quiet near the middle of a high pressure system. Wind is the passage of air from a high pressure location to a low pressure area.... So essentially air is always moving. Weathering and erosion are caused by wind. Weathering is caused by wind blowing debris against cliffs and huge rocks. This wears down the rock, reducing it to sand and dust. Sand and dust are also eroded by wind. 2. Rocks are tough and durable, but they don't last forever. Weathering and erosion are processes that occur as a result of forces such as wind and water breaking down rocks. Weathering is the process through which rocks deteriorate. Weathering is caused by a variety of factors, including climate change.

Deposition occurs when wind dies down or stops blowing happens.

Deposition refers to the depositing of sediment carried by wind, flowing water, the sea, or ice. Sediment can be carried in the form of pebbles, sand, and mud, or as salts dissolved in water. Organic activity may later deposit salts.

The fractured rock fragments or soil particles are accumulated in an environment where the energy of the flowing river is reduced. It settles the particles, which compact and form sedimentary rocks over time.

Deposition can occur when a river enters a shallow area of water or when the volume of water decreases, such as after a flood or during a drought. Deposition is common at the mouth of a river near the end of its journey.

Thus, the answer is deposition.

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