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vazorg [7]
3 years ago
8

Help

Physics
2 answers:
Ede4ka [16]3 years ago
7 0

<h2>sin (x + y) ≈ 0.58</h2>

<h3>Further explanation</h3>

Firstly , let us learn about trigonometry in mathematics.

Suppose the ΔABC is a right triangle and ∠A is 90°.

<h3>sin ∠A = opposite / hypotenuse</h3><h3>cos ∠A = adjacent / hypotenuse</h3><h3>tan ∠A = opposite / adjacent </h3>

There are several trigonometric identities that need to be recalled, i.e.

cosec ~ A = \frac{1}{sin ~ A}

sec ~ A = \frac{1}{cos ~ A}

cot ~ A = \frac{1}{tan ~ A}

tan ~ A = \frac{sin ~ A}{cos ~ A}

Let us now tackle the problem!

If sec (y) = 25/24 , then we can assume that :

adjacent side = 24 cm

hypotenuse = 25 cm

opposite side = \sqrt{25^2 - 24^2} = 7 ~ cm

sin (y) = opposite / hypotenuse = 7/25

If sin (x) = 1/3 , then we can assume that :

opposite side = 1 cm

hypotenuse = 3 cm

adjacent side = \sqrt{3^2 - 1^2} = \sqrt{8} ~ cm

cos (x) = adjacent / hypotenuse = \sqrt{8}/3 = \frac{2\sqrt{2}}{3}

\sin (x + y) = \sin x ~ \cos y + \cos x ~ \sin y

\sin (x + y) = \frac{1}{3} \times \frac{24}{25} + \frac{2\sqrt{2}}{3} \times \frac{7}{25}

\large {\boxed {\sin (x + y) = \frac{24 + 14\sqrt{2}}{75} \approx 0.58} }

<h3>Learn more</h3>
  • Calculate Angle in Triangle : brainly.com/question/12438587
  • Periodic Functions and Trigonometry : brainly.com/question/9718382
  • Trigonometry Formula : brainly.com/question/12668178

<h3>Answer details</h3>

Grade: College

Subject: Mathematics

Chapter: Trigonometry

Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse , Triangle , Fraction , Lowest , Function , Angle

Blizzard [7]3 years ago
6 0
We can approach this in another way.
We know that sin(∅) = height / hypotenuse.

Thus, for x, height is 1 and hypotenuse is 3. Using Pythagoras theorem,
3² = 1² + b²
b = √8
cos(x) = b/hypotenuse
cos(x) = √8 / 3

Now, lets consider y:
sec(y) = 1 / cos(y) = 1 / base / hypotenuse = hypotenuse / base
The hypotenuse is 25 and the base is 24. We again apply Pythagoras theorem to find the third side, which works out to be:
height = 7
sin(y) = height / hypotenuse
sin(y) = 7/25

Now, sin(x + y) =
sin(x)cos(y) + sin(y)cos(x)
= (1/3)(24/25) + (√8 / 3)(7/25)
= 8/25 + 7√8/75
= (24 + 14√2) / 75
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My mass is 65 kg and on Earth this equals a weight of 640 N, but on the moon where gravity is 1.7 m/s² my
Helen [10]

Your weight on the moon given the data from the question is 110.5 N

<h3>Definition of mass and weight </h3>

Mass is simply defined as the quantity of matter present in an object. The mass of an object is constant irrespective of the location of the object.

Weight is simply defined as the gravitational pull on an object. The weight of an object varies from place to place due to gravity.

<h3>Relationship between mass and weight </h3>

Mass and weight are related according to the following equation

Weight (W) = mass (m) × Acceleration due to gravity (g)

<h3>How to determine the weight on the moon</h3>
  • Mass (m) = 65 Kg
  • Acceleration due to gravity on the moon (g) = 1.7 m/s²
  • Weight (W) =?

W = mg

W = 65 × 1.7

W = 110.5 N

Learn more about mass and weight:

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What will be the change in velocity of a 850kg car if a force of 50,000 N
yKpoI14uk [10]

Answer:

29.412m/s

Explanation:

F=ma where F= force, m= mass, and a=acceleration

we also know that,

a = Δv / t where Δv = change in velocity and t = time

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Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
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