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4 years ago

Which of the following happens to an object in uniform circular motion?

Physics

1 answer:

podryga [215]4 years ago

5 0

Answer:

As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

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A sewing machine uses 1.6kWh of electricity in one day. If electricity costs 9p per unit, what is the total cost in pence of usi

atroni [7]

Answer:

$\huge\boxed{\sf 1.6\ units = 14.4p}$

Explanation:

Since,

So,

1.6 kWh = 1.6 units

If,

1.6 units = 9p × 1.6

1.6 units = 14.4p

$\rule[225]{225}{2}$

8 0

2 years ago

In which direction does a convergent boundary move?

VladimirAG [237]

Answer:

Toward each other teehee merry christmas

Explanation:

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3 years ago

Read 2 more answers

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

snow_lady [41]

Answer:

<em>The frequency of of the note = 131 Hz.</em>

Explanation:

<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

<em>Given: v = 343 m/s, λ = 2.62 m.</em>

<em>Substituting these values into equation 2</em>

<em>f = 131 Hz</em>

<em>Thus the frequency of of the note = 131 Hz.</em>

8 0

3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::