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Naddik [55]
1 year ago
5

Scenario 3Starting at rest, a 3 kg ball is dropped from the side of a bridge and strikes the ground below at 35m/s. What is the

height of the bridge?
Physics
1 answer:
Nonamiya [84]1 year ago
4 0

   The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.

<h3>How can the height of a dropped ball be determined?</h3>

    Y = 1/2 g t 2, where y is the height above the ground, g = 9.8 m/s2, and t = 1.3 s, is the formula for problems like these. Any freely falling body with an initial velocity of zero meters per second can use this formula. figuring out how much y is.

   A ball drops from the top of a building and picks up speed as it descends. Its speed is increasing by 10 m/s every second. What we refer to as motion with constant acceleration is, for example, a ball falling due to gravity.

    The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.

To Learn more About potential energy, Refer:

brainly.com/question/14427111

#SPJ10

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A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch p
kirill115 [55]

Answer:

Change in mechanical energy, \Delta E=283.2\ J

Explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

K=\dfrac{1}{2}mv^2

K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

P=12\ kg\times 9.8\ m/s^2\times 18\ m

P =2116.8 J

The change in mechanical energy is given by :

\Delta E=K-P

\Delta E=2400-2116.8

\Delta E=283.2\ J

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.

8 0
3 years ago
Which of the following changes will increase the frequency of an oscillating pendulum?
svlad2 [7]

Answer:

c. an increase in the length of the rope.

6 0
2 years ago
How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The
ra1l [238]

Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE* = 0,

KE* = 1/(2mv^2)

Now since

PE + KE = Total energy =PE* + KE*

Therefore,

qV + 0 = 0 + 1/2mv^2

Or

KE = qV = 1.6 10^-19 J

4 0
3 years ago
Long sand ridges oriented at right angles to the wind are called _____ dunes
S_A_V [24]

The answer is:

Transverse dune

The explanation:

Transverse dune : is abundant barchan dunes It  may merge into barchanoid ridges, which then grade into linear .

The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.

It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.

Transverse dunes have a gently sloping windward side and a steeply sloping leeward side.

They general form in areas of sparse vegetation and abundant sand are transverse dunes.

3 0
3 years ago
Read 2 more answers
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
3 years ago
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