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djverab [1.8K]
3 years ago
5

A jet travels 2250 km westward at a speed of 960 km/hr. it then encounters a strong headwind and its speed drops to 805 km/hr fo

r the next 1320 km of its westward trip. what is the average velocity for the trip?
Physics
1 answer:
Kipish [7]3 years ago
4 0
For the first part of the trip:
we have velocity = 960 km/hr and distance = 2250 km
velocity = distance / time
time1 = distance / velocity = 2250 / 960 = 2.34375 hours

For the second part of the trip:
we have velocity = 805 km/hr and distance = 1320 km
velocity = distance / time
time2 = distance / velocity = 1320 / 805 = 1.63975 hours

For the total trip:
we have:
total distance covered = 2250 + 1320 = 3570 km
total time taken = 2.34375 + 1.63975 = 3.9835 hours
average velocity = total distance / total time
average velocity = 3570 / 3.9835 = 896.1968 km/hr
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The separation between the slits is d = 8.96

What is fringe width?

  • Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
  • Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d   = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 )  = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428  = 6.222  

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d   = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 )  = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714  = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8.  The ninth one will not be seen since the screen is less a little less in width.

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<u>The complete question is -</u>

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0
1 year ago
A chain of length L has a mass of m which is uniformly distributed. When it is placed on a smooth horizontal table, 1/4 of its l
krok68 [10]

Gravitational potential energy has changed from the beginning to the chain just sliding away from the table willl be MgL/24.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

Length of hanging part, L/4

The gravitational potential energy is;

Work = Gravitational potential energy

Work done =force×displacement

\rm W = \frac{Mg}{3} \times \frac{L}{8} \\\\ W = \frac{MgL}{24}

Hence, gravitational potential energy has changed from the beginning to the chain just sliding away from the table willl be MgL/24.

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6 0
2 years ago
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at
VARVARA [1.3K]

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

6 0
3 years ago
If a force of 10 n is applied to an object with a mass of 1kg the object will accelerate at
givi [52]
We Know, F = m*a
Here, F = 10 N
m = 1 Kg

Substitute their values in the equation,
10 = 1 * a
a = 10/1
a = 10

So, your final answer & the acceleration of the object would be 10 m/s²

Hope this helps!
3 0
3 years ago
NEED HELP ASAP!
fiasKO [112]
The answer to this question is d because prejudices is not based on experience or reasons
3 0
3 years ago
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