An airplane flies 33 m/s due east while experiencing a tailwind

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago

How can i calculate distances between objects by using the concepts of echo location

dangina [55]

Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return

5 0

3 years ago

Explain the phenomenon of polarization when a dielectric slab is subjected to an electric<br> field?

Oxana [17]

A dielectric, insulating material, or an extremely bad conductor of electrical current. Due to the absence of loosely bound, or free, electrons that could wander through the material, unlike metals, dielectrics practically do not conduct current when exposed to an electric field. Electric polarization takes place instead.

<h3>What is an Electric field?</h3>

An electric field is an electrical property associated with every point in the space of any form of charge. An electric field is also described as the electric force per unit charge.
Variable magnetic fields or electric charges are frequently the cause of electric fields. Volts per meter, a unit used in the SI, express electric field strength.
The force acting on the positive charge is assumed to be exerted in the direction of the field. The electric field is directed radially inwards toward the negative point charge and radially outwards from the positive charge.
Electric charge or magnetic fields with variable amplitudes can produce an electric field. The attraction forces that keep together atomic nuclei and electrons at the atomic scale are brought on by the electric field.

The phenomenon of polarization when a dielectric slab is subjected to an electric field:

A dielectric, insulating material, or an extremely bad conductor of electrical current. Due to the absence of loosely bound, or free, electrons that could wander through the material, unlike metals, dielectrics practically do not conduct current when exposed to an electric field. Electric polarization takes place instead.

To learn more about the electric field, refer to:

brainly.com/question/14372859

#SPJ9

8 0

2 years ago

Special relativity can be used to study an object in which frame of reference?

Tresset [83]

Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.

3 0

3 years ago

Read 2 more answers

Will mark as brainliest if correct!!!!!!!

lutik1710 [3]

C. A could be ruby (speed of light = 170,000 km/s); B could be diamond (speed of light = 120,000 km/s).

Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$ (1)

where

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

$n=\frac{c}{v}$

Using this definition, we can rewrite eq.(1) as

$\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}$

Where $v_2$ is the speed of light in the 2nd medium and $v_1$ the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

$\theta_2 < \theta_1$

This means that

$\frac{\sin \theta_2}{\sin \theta_1}$

And therefore,

$\frac{v_2}{v_1}$

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0

3 years ago