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Triss [41]
1 year ago
13

During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying

in your chaise lounge soaking up the warm Mediterranean sun, a large glob of seagull poop hits you in the face. Since you got an “A” in ICPE you are able to estimate the impact velocity at 98.5 m/s. Neglecting air resistance, calculate how high up the seagull was flying when it pooped.
Physics
1 answer:
adelina 88 [10]1 year ago
6 0

The maximum height attained is 460 m.

<h3>What is the maximum height?</h3>

We know that the final velocity of a body is 0 m/s at the maximum height which is the greatest height that is attained by the body. We now use the formula;

v^2 = u^2 -2gh

Given that v = 0 m/s

u^2 = 2gh

h = u^2/2g

v = final velocity

u = initial velocity

h = maximin height

g = acceleration due to gravity

h = (95)^2/2 * 9.8

h = 460 m

Learn more about maximum height:brainly.com/question/6261898

#SPJ1

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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0
3 years ago
A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
3 years ago
A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H
IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0
3 years ago
A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be
makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , \omega =8.5\, \frac{rad}{sec}

Mass m is moving on platform in a circle with radius , r=0.20\, m

Force sensor reading to which spring is attached , F=4.8\, N

Now for the mass m to move in circle the required centripetal force is given by F=m\omega ^{2}r

=>4.8=m\times 8.5 ^{2}\times 0.20

=>m=0.332\, kg

Thus the mass m is 0.332 kg or 332 gm

7 0
3 years ago
Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5
lesantik [10]

Answer:

The telescope can gather light 1600 times more than the human eyes can!

Explanation:

The light gathering ability of an optical element is directly proportional to its area of opening.

So, in comparing the light gathering abilities for two objects, it is just the ratio of their area of opening.

Let the diameter of the telescope be D = 1 m

And the diameter of the human eyes be d = 2.5 cm = 0.025 m

Light gathering ability of the telescope compared to the eyes = D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

The telescope can gather light 1600 times more than the human eyes can!

Hope this Helps!!!

7 0
3 years ago
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