Answer:
### Final answer:

### Explanation:

### Learn more about Mechanics and Energy Conservation here:

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.

**"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.**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.**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.**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.

#SPJ12

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?Express your answer in micrometers(not in nanometers).

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

(b) Calculate the magnification.

**Answer:**

A) Calculate the distance

Answer:

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.

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.

- 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).
- 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**.

#SPJ3

**Answer:13.6 cm**

**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.

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**.

#SPJ3

**Answer:**

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**

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.

To learn more about **gravitational force**, here

#SPJ2

**Answer:**

THE ANSER IS B

**Explanation:**

(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

(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

**(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