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lutik1710 [3]
3 years ago
6

Jasmine is diving off a 3-meter springboard. her height in meters above the water when she is x meters horizontally from the end

of the board is approximated by the equation h=-x^2+3x+3. what is the maximum height jasmine will reach on her dive?
Physics
1 answer:
olganol [36]3 years ago
8 0

Answer:

5.25 m

Explanation:

Given;

The height equation h;

h=-x^2+3x+3

Where;

h = the height above water

x = horizontal distance from the end of the board

The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

2x = 3

x = 3/2

Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

h max = 21/4 = 5.25 m

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N which order did the events forming our solar system occur?
brilliants [131]

Answer:

The solar nebula became hot and dense because of that it pulling in more gas. This flattened into a rotating disk. It  spun  faster and faster, forming the Sun.

Explanation:

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4 0
3 years ago
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Two forces,one of 12 N and another of 24 N,act on a body in such a way that they make an angle of 90degree with each other.Find
jok3333 [9.3K]

Answer:

26.83 N.

Explanation:

If the angle between two vector is 90°, to get the resultant, we use Pythagoras theorem.

a² = b²+c²......................... Equation 1

Where a = R = Resultant, b =  12 N, c = 24 N.

Substitute these values into equation 1

R² = 12²+24²

R² = 144+576

R² = 720

√R² = √720

R = 26.83 N.

Hence, the result of the two force is 26.83 N.

4 0
3 years ago
A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n
andreyandreev [35.5K]

Answer:

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s

7 0
3 years ago
A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg
Evgen [1.6K]

Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

=58.96Hz

Let, n th harmonic is the hightest frequency, then

(58.93)n = 20000

∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

learn more about frequency refer:

brainly.com/question/254161

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7 0
2 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
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