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tekilochka [14]

2 years ago

10

A ship is moving at a speed of 25 km/h parallel to a straight shoreline. The ship is 9 km from shore and it passes a lighthouse

at noon. (a) Express the distance s between the lighthouse and the ship as a function of d, the distance the ship has traveled since noon. s = f(d) = (b) Express d as a function of t, the time elapsed since noon. d = g(t) = (c) Find f ∘ g. (f ∘ g)(t) =

1 answer:

sp2606 [1]2 years ago

6 0

Explanation:

It is given that,

Speed of the ship, v = 25 km/h

Distance between the ship and the shore is 9 km

Let s is the distance between the lighthouse and the ship and d is the distance traveled by the ship since noon.

(a) Using Pythagoras theorem in the attached diagram as :

$s^2=9^2+d^2$

$s^2=81+d^2$

$s=\sqrt{81+d^2}$

or $f(d)=\sqrt{81+d^2}$

(b) Let g(t) is the function at t = 0 and at 12 pm.

$f(g)=fog$

Since, distance, d = 25t

So, $f(g(t))=\sqrt{9^2+(25t)^2}$

$f(g(t))=\sqrt{81+(652t)^2}$

Hence, this is the required solution.

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A professional boxer hits his opponent with a 1025 N horizontal blow that lasts 0.150 s. The opponent's total body mass is 116 k

mylen [45]

Answer:

The impulse is $I = 153.8 \ N \cdot s$

The opponents velocity is $v= 1.33 m/s$

The opponents head recoils velocity $v_r = 30.8 \ m/s$

Explanation:

From the question we are told that

The force of the blow is $F= 1025 \ N$

The duration of the blow is $t = 0.150$

The mass of the opponent is $m_o = 116 \ kg$

The mass of the opponents head is $m_h = 5 \ kg$

The impulse the boxer imparts is mathematically represented as

$I = F * t$

substituting values

$I = 1025 * 0.150$

$I = 153.8 \ N \cdot s$

The impulse can also be mathematically evaluated as

$I = m_o * v$

substituting values

$153.8 = 116 * v$

$v= \frac{153.8}{116}$

$v= 1.33 m/s$

The recoil velocity is mathematically represented as

$v_r = \frac{I}{m_h}$

substituting values

$v_r = \frac{153.8}{5}$

$v_r = 30.8 \ m/s$

5 0

2 years ago

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

3 0

2 years ago

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
<u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
<u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

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2 years ago

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

8 0

2 years ago

A cart moves along a track at a velocity of 3.5 cm/s. when a force is applied to the cart, its velocity increases to 8.2 cm/s. i

oksian1 [2.3K]

Acceleration, in physics, is the rate of change of velocity of an object with respect to time. <span> Anytime an object's velocity is changing, the object is said to be </span><span>accelerating. It can be calculated as follows:

acceleration = 8.2 - 3.5 / 1.5 = 3.1 m/s</span>²

Hope this answers the question.

5 0

2 years ago

Read 2 more answers