You are standing on the edge of a ravine that is 17.5 m wide. You notice a cave on the opposite wall whose ceiling is 7.3 m below your feet. The cave is 4.2 m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang? The acceleration of gravity is 9.8 m/s 2 .

Answers

Answer 1
Answer:

Answer:

v₀ₓ = 14.34 m / s

Explanation:

We can solve this problem using the projectile launch equations.

Let's look for the time it takes to descend to the height of the cave

    y = v_(oy) t - ½ g t²

As it rises horizontally the initial vertical speed is zero  

    y = 0 - ½ gt²

    t = √2 y / g

    t = √2 7.3 / 9-8

    t = 1.22 s

This is the same time to cross the ravine

    x = v₀ₓ t

    v₀ₓ = x / t

    v₀ₓ = 17.5 / 1.22

    v₀ₓ = 14.34 m / s

This is the minimum speed.


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A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

Answers

The reflected beam experienced a phase change of about 180°.

What is reflection in the glass surface?

According to Snell's law, the light that incident on the glass surface will be reflected and transmitted at an angle equals to the angle of incidence.  

By the observation of refractive index of the glass for the normal incidence only 4% of the light is transmitted or reflected.

The light passing through glass is not only reflected on the front surface, but also on the back. For several times the light will gets reflected back and forth. So, the total reflectance through a glass window can be calculated as

                                    2·R / (1+R).

Thus, A light wave travelling in air is reflected by a glass barrier will undergo a  phase change of 180°, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air.

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Answer:

180 degree phase change

Explanation:

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

P_r=(IA)/(c)

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

P=140W=140(J)/(s)\n0.4*140J=56J

56J is the electromagnetic energy.

The area of the bulb is

A_b=4\pi r^2=4\pi (14m)^2=2463m^2

The radiation pressure is

P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa

hope this helps!!

Which of the following organisms has an adaptation that will allow it to survive in tundra biome? *A.)Plants with roots that are short and grows sideways with hairy stems and small leaves.
B.)Plants that have broad leaves to capture sunlight and long roots to penetrate the soil.
C.)Animals with thin fur that allows them to get rid of heat efficiently.
D.)Animals with long tongues for capturing prey and sticky pads for climbing trees.

Answers

Answer:

the awnser is A becuse the hair help.

We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.

Answers

Answer:

* roller skates and ice skates.

* roller coaster

Explanation:

One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.

Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.

A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?

Answers

Answer:

The number is  N  = 300

Explanation:

From the question we are told that

   The  net charge is  Q =  -4.8 *10^(-17 ) \  C

Generally the charge on a electron is e = - 1.60 *10^(-19 ) \ C

Generally the number of excess electrons is mathematically represented as

      N  =  (Q)/(e)

=>  N  =  (-4.8 *10^(-17))/(-1.60 *10^(-19))

=>  N  = 300

A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

Answers

Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

Explanation:

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

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