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3 years ago

15

An object has an angular velocity of 1.0 rad and an angular acceleration of 0.5 rad/s. Is the speed of its rotation increasing o

r decreasing?

1 answer:

alexdok [17]3 years ago

6 0

Answer:

increasing

Explanation:

Given that,

Initial angular speed =1 rad/s

Angular acceleration = 0.5 rad/s²

We know that the angular acceleration is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

Where

$\omega_f$ city and t is time is the final angualr velo

As the angular acceleration of the object is positive. It means that the final speed of the object is more than that initial. As a result, the speed of its rotation is increasing.

You might be interested in

A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.

Marizza181 [45]

For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2

5 0

3 years ago

In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,

taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

$f = \frac{1}{2\pi \sqrt{LC} }$

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

$f' = \frac{1}{2\pi \sqrt{2L2C} }$

$f' = \frac{1}{2\pi \sqrt{4LC} }$

$f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}$

$f' = \frac{1}{2} f$

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

#SPJ4

6 0

2 years ago

I need help on all of this

wel

dude is -2m/s, ...ettte is +2m/s

pos vel is when dudette, eg, is going in increasing x, in this case.

neg vel is when dudette, eg, is going in decreasing x, in this case. ie she turns round and runs

zero vel is zero speed. dudette standing still

positive vel neg pos top dia

no

yes

yes

yes

yes

yes

all yes looks like ...

4 0

3 years ago

Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is t

NikAS [45]

Answer:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

$\Delta x = (2n + 1)\frac{\lambda}{2}$

given that path difference from two loud speakers is given as

$\Delta x = 5.80 m - 3.90 m$

$\Delta x = 1.90 m$

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

$\Delta x = 1.90 = \frac{7\lambda}{2}$

$\lambda = \frac{2 \times 1.90}{7}$

$\lambda = 0.54 m$

now for frequency we know that

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.54} = 632 Hz$

7 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago