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Citrus2011 [14]
2 years ago
5

1. If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released. It will oscillate. I

f is displaced 0.120m from its equilibrium position and released with zero initial speed, then after 0.800s its displacement is found to be 0.120m on the opposite side, and it has passed the equilibrium position once during this interval. Find, a) the amplitude; b) the period; c) the frequency.
Physics
1 answer:
AURORKA [14]2 years ago
3 0

Answer: a) 0.12m; b) 1,6 s; c) 0.625 1/s

Explanation: The simple harmonic movement can be described by a sin or cosine function  in time.

This can be in the form:

X(t)= A Sin/Cos (wt+φ) where φ is initial phase o position at t=0

w the angular frequency are related to the frequency (f) as 2Pif

and f=1/T period of oscillating

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Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th
Citrus2011 [14]

Answer:

(a) Rm = 268.4 m

(b) f = 6

Explanation:

The horizontal range of a projectile is given by the following formula:

R = V₀² Sin 2θ/g

(a)

For moon:

R = Range on moon = Rm

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on moon = (9.8 m/s²)/6 = 1.63 m/s²

Therefore,

Rm = (28 m/s)²Sin (2*17°)/(1.63 m/s²)

<u>Rm = 268.4 m</u>

(b)

For Earth:

R = Range on Earth = Re

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

Re = (28 m/s)²Sin (2*17°)/(9.8 m/s²)

Re = 44.7 m

Therefore.

f = Rm/Re = 268.4 m/44.7 m

<u>f = 6</u>

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A Styrofoam slab has thickness h and density ρs. When a swimmer of mass m is resting on it, the slab floats in fresh water with
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Waves hitting at an angle and then bending around features of the coast is known as
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<span>Waves hitting at an angle and then bending around features of the coast is known as Wave refraction
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2 years ago
The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
2 years ago
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