Someone please help me with this question

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

DanielleElmas [232]

Answer:

$\alpha =-2.2669642\times^{-10}rad/s^2$

Explanation:

Angular acceleration is defined by $\alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}$

Angular velocity is related to the period by $\omega=\frac{2\pi}{T}$

Putting all together:

$\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})$

Taking our initial (i) point now and our final (f) point one year later, we would have:

$\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s$

$T_i=0.0786s$

$T_f=0.0786s+7.03\times10^{-6}s$

So for our values we have:

$\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2$

Where the minus sign indicates it is decelerating.

8 0

3 years ago

Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. The

goldfiish [28.3K]

I think the answer to your question is B

7 0

3 years ago

A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force

Vlada [557]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec
vf = 0 m/sec 
Fμ = 7100 N (Force due to friction) 
Fg = 14700 N 
With the force due to gravity, you can find the mass of the car: 
F = ma 
14700 N = m (9.8 m/sec²) 
m = 1500 kg 
Now, we can use the equation again to find the deacceleration due to friction: 
F = ma 
7100 N = (1500 kg) a 
a = 4.73333333333 m/sec² 
And now, we can use a velocity formula to find the distance traveled: 
vf² = vo² + 2a∆d 
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d 
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d 
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d 
∆d = 66.0211267605634 m 
∆d = 66.02 m

7 0

3 years ago

After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter

Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, $\Delta L=3.1\ cm$

(a) The change in length per unit original length is called strain. So,

$\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193$

(b) The force acting per unit area is called stress.

$\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa$

(c) The ratio of stress to the strain is called Young's modulus. So,

$Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2$

Hence, this is the required solution.

8 0

2 years ago

What tension would you need to make a middle c (261.6 hz) fundamental mode on a 1 m string (for example, on a harp)? the linear

Allisa [31]

The frequency of middle C on a string is
f = 261.6 Hz.

The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m

The length of the string is L = 1 m.

Let T = the tension in the string (N).
The velocity of the standing wave is
$v= \sqrt{ \frac{T}{\rho} }$

In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
$\sqrt{ \frac{T}{\rho} } =f \lambda = fL \\\\ \frac{T}{\rho} = (fL)^{2} \\\\ T = \rho (fL)^{2}$

From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N

Answer: 136.9 N (nearest tenth)

4 0

3 years ago