An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

1. "Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
2. Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
3. Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
4. Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

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Related Questions

A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.

(b) Calculate the magnification.

A) Calculate the distance

Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit student ID. (Example: For 20245347, A = 47, B = 347, and C = 19) A train moves at an average speed of (23.0 + A) m/s for (250.0 + B) seconds and then at an average speed of (45.0 + C) m/s for (800.0 + B) seconds. Determine the average speed for the entire time in meters per second (m/s). Round your final answer to 3 significant figures.

66.053m/s

Explanation:

A = 47

B = 347

C = 19

Train moves at

(23 + A)m/s

= 23 + 47 = 60m/s

At (250.0+B) seconds

250.0+347 =

547 seconds

Distance d,

= 70 x 597

= 41790

It also moves at

(45.0 + c)

= 45 + 19

= 64m/s

Time = 800 + B

= 800 + 347

= 1147

Distance,

= 64 x 1147

= 73408m

Total distance,

= 73408 + 41790

= 115,198

Total time,

= 597 + 1147

= 1744

Average speed,

= Total distance / total time

= 115198/1174

= 66.053m/s

The average speed over the entire time can be calculated by first finding the distances the train travels over both periods, then finding the total distance and the total time, and finally dividing the total distance by the total time. The value must be rounded to three significant figures.

Explanation:

You can find the average speed of the train over the full-time interval by dividing the total distance travelled by the total time. To begin with, you would have to find the distances the train covered during both periods.

1. The distance (D1) it travelled during the first period can be found by multiplying the average speed (23.0 + A) by the time (250.0 + B).
2. The distance (D2) it travelled during the second period can be calculated by multiplying the average speed (45.0 + C) by the time (800.0 + B).

Then you add D1 and D2 to get the total distance (TD). This will be (D1 + D2). The total time (TT) will be found by adding both time intervals, which means it equals (250.0 + B) + (800.0 + B). You then divide the total distance by the total time to get the average speed, i.e., TD/TT. Lastly, round the average speed to 3 significant figures.

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A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Explanation:

Given

v(image distance)=-8.5 m

height of object=6 mm

height of image =37.5 cm

and magnification of concave mirror is given by

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = = 0.00453 m

Therefore,

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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THE ANSER IS B

Explanation:

The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

(1) The acceleration of the car will be

(2) The time taken

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

Here

u= 14

(2) The time is taken for the same acceleration to 20

u=20

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

v=10

u=20

Thus

(1) The acceleration of the car will be

(2) The time taken

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s

To know more about the Equation of the motion follow

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(a)

The car's acceleration is given by

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

where in this case we have

is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

(c)

As before, we can use the equation

Here we have

is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find