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Fed [463]
3 years ago
7

answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th

at oscillates in simple harmonic motion. Calculate the period of motion.
Physics
2 answers:
victus00 [196]3 years ago
8 0

Answer:

0.695s

Explanation:

From Hooke's law, the restoring force is given has

F = -ky .......1

Where F is the force, y is the spring displacement and k force constant of the spring.

Also recall,

F=mg ............ 2

Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

Given that g=9.8m/s2 , y is 3.4cm and g is 8g

K×3.4/100m =8/1000kg × 9.8m/s2

K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

K= 0.0784÷0.035

K=2.24N/m

Mass ofvthe second object is 25g =0.025kg

Period of oscillation T

T=2π√m/k

T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

notsponge [240]3 years ago
8 0

Answer:

df,jgukilhugyftdrswjdfggfjgjfjgfthf

Explanation:

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A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reach
natka813 [3]

The total number of revolutions made by the wheel is closest to is 28.2 revolutions. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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3 years ago
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An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How
Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

g=24 m/s^2

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

s=0+\frac{1}{2}(24)(1)^2=12 m

6 0
3 years ago
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
Which of the following is the equation for calculating daily caloric needs? A. BMR calories + Digestion calories B. BMR calories
labwork [276]

Explanation:

To determine your total daily calorie needs, multiply your BMR by the appropriate activity factor, as follows: If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2. If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.

<em><u>HAPPY TO HELP</u></em>

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