Which are the components of a typical refracting telescope? hints?

A pendulum is swinging next to a wall. The distance from the bob of the pendulum to the wall varies in a periodic way that can b

nexus9112 [7]

The formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds is

H(t) = 15 -6sin(2.5π(t -0.5))

Detailed explanation:

The function can be expressed as the following for the midline M, amplitude A, period T, and time t0 at which the function deviates from the midline:

H(t) = M -Asin(2π/T(t -t0))

The equation is based on the parameters M=15, A=6, T=0.8, and t0 = 0.5.

H(t) is equal to 15 -6sin(2.5π(t -0.5))

<h3>What is function?</h3>

The trigonometric functions in mathematics are real functions that connect the right-angled triangle's angle to the ratios of its two side lengths .In all areas of study that involve geometry, such as geodesy, solid mechanics, celestial mechanics, and many others, they are widely used.

To learn more about functions visit:

brainly.com/question/15607563

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The correct question is:

A pendulum is swinging next to a wall. The distance from the bob of the swinging pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function.

The function has period 0.80.80, point, 8 seconds, amplitude 6 \text{ cm}6 cm6, start text, space, c, m, end text, and midline H = 15 \text{ cm}H=15 cmH, equals, 15, start text, space, c, m, end text. At time t = 0.5t=0.5t, equals, 0, point, 5 seconds, the bob is at its midline, moving towards the wall.

Find the formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds. Define the function using radians.

5 0

1 year ago

a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.

inessss [21]

Answer:

B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

B = $- \frac{P}{ \frac{\Delta V}{V} }$

The negative fate is for the module to be positive since the volume change is negative

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

B = $- \frac{10^4}{ \frac{-0.20}{5000} }$ = $\frac{10^4}{4 \ 10^{-5} }$

B = 2.5 10⁸ Pa

4 0

3 years ago

A 15 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 53

blagie [28]

Answer:

0.0613°C

Explanation:

the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s

the bullet moves with different kinetic energies before and after the penetration, therefore

Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)

= $\frac{1}{2}$ × 15×10⁻³ × (865² - 534²)

= 3.47 × 10⁻³J

this loss in energy is transferred to the water, therefore

change in temperature = $\frac{Q}{m C}$

where c = heat capacity of water = 4.19 x 10^3

m = mass of water = 13.5 kg

= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }

=0.0613°C

5 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$