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Margaret [11]
2 years ago
12

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the

drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.60 m, and x = 7.1 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.
Physics
1 answer:
Ivahew [28]2 years ago
6 0

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

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The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many
Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0
2 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
yuradex [85]

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

     V=4x8=32V

k.e=eV= 2X32=64eV

3 0
2 years ago
This is really urgent
hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is n=\frac{c}{v}

26) Critical angle: [tex]sin \theta_c = \frac{n_2}{n_1}[/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

with

n_1, n_2 are the refractive index of 1st and 2nd medium

\theta_1, \theta_2 are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

  • A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
  • A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

  • A material with refractive index of 1.1
  • A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1 (1)

For light moving from air to water:

n_1 \sim 1.00 is the index of refraction of air

n_2 = 1.33 is the index of refraction ofwater

In this case, \frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75

For light moving from air to glass,

n_2 = 1.51 is the index of refraction of glass

And so

\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66

From eq.(1), we see that the angle of refraction \theta_2 is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

n=\frac{c}{v}

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

  • The higher the index of refraction, the slower the light
  • The lower the index of refraction, the faster the light

26)

From Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, n_1 > n_2, so \frac{n_1}{n_2}>1, and since sin \theta_2 cannot be larger than 1, there exists a maximum value of the angle of incidence \theta_c (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

sin \theta_c = \frac{n_2}{n_1}

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

sin \theta_c = \frac{n_2}{n_1}

For the interface glass-air,

n_1 \sim 1.51\\n_2 = 1.00

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}

For the interface glass-water,

n_1 \sim 1.51\\n_2 = 1.33

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0
2 years ago
Seema knows the mass of basketball. What other information is needed to find the balls potential energy
Lelu [443]

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field.  In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.  

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy U will be:  

U=mgh  

Where:

m is the mass of the ball

g=9.8 m/s^{2} is the acceleration due gravity (assuming the ball is on the Earth surface)

h is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

8 0
2 years ago
Read 2 more answers
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