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agasfer [191]
3 years ago
10

Define Simple harmonic motion (SHM). A body moving with SHM has an amplitude of 10 cm and a frequency of 100 Hz. Find (a) the pe

riod of oscillation (b) the acceleration at the maximum displacement (c) the velocity at the centre of motion.
Physics
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

Explanation:

Simple harmonic motion is the motion that involves the cyclic and periodic motion of an object about a central point or equilibrium position, due to the effect of a restoring force, such that the object has the same maximum displacement during its motion on either side of the equilibrium point

(a) The parameters of the body moving with SHM are;

The amplitude of the motion of the body, a = 10 cm

The frequency of the complete cycles of the motion of the body, f = 100 Hz

The period of the oscillation, 'T', is the time it takes o complete one cycle

T = 1/f

∴ T = 1/100 Hz = 0.01 seconds

(b) The acceleration, a = -A·ω²·sin(ω·t)

Where;

One complete oscillation is given as ω·t = 2·π

Therefore at either end at the maximum displacement, which is at either end of the equilibrium position is reached at half a complete cycle, or ω·t = 2·π/2 = π

At the maximum displacement a = -A·ω²·sin(ω·t) = A·ω²·sin(π) = 0

Therefore, we have;

The acceleration, of the body at the maximum displacement, a = 0 m/s².

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What is the mass of 2000 ml of an intravenous glucose solution with a density of 1.15 g/ml?
Scorpion4ik [409]

According to the following formula, the answer is 2,300 g or 2.3 kg:

Volume (m)/Mass (m) Equals Density (p) (V)

Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:

p=m/V

1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.

2,000 mL of x g = 1.15 g of g/mL

2.3 kg or 2,300 g for x g.

<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>

Its active ingredient is glucose. This medication includes 50 g of glucose per 1000 ml (equivalent to 55 g glucose monohydrate). 50 mg of glucose is present in 1 ml (equivalent to 55 mg glucose monohydrate). A transparent, nearly colourless solution of glucose in water is what is used in glucose intravenous infusion (BP) at 5% weight-to-volume.

Patients who are dehydrated or who have low blood sugar levels get glucose intravenously. Other medications may be diluted with glucose intravenous infusion before being injected into the body. Other diseases and disorders not covered above may also be treated with it.

learn more about  glucose intravenous infusion refer

brainly.com/question/7057736

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5 0
1 year ago
Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c
lana [24]

Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

8 0
3 years ago
A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

F = m a\\\\a =\frac{F}{m}

We use movies and find lips

\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t

The moment is defined by

\to p = m v

The moment change

\Delta p = m v - m v_0

Let's replace the speeds in this equation

\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0  + F t - m v_0\\\\\Delta p = F t

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0
3 years ago
g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pasca
Masja [62]

Answer:

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

3 0
3 years ago
Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance b
AveGali [126]

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m

To produce destructive interference at a listener standing in front of them,

d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
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