All categories

2 years ago

Why does the spectroscopic parallax method only work for main sequence stars?.

Physics

1 answer:

Brums [2.3K]2 years ago

5 0

Answer:

Only main sequence stars have a well-defined relationship between spectral type and luminosity.

Explanation:

Low-mass stars have much longer lifetimes than high-mass stars.

You might be interested in

Which statement about an atom is correct?(1 point)

alexdok [17]

Answer:

The electron has a negative charge and is found outside of the nucleus.

Explanation:

1.) Neutrons have no charge, not a negative charge.

2.) Neutrons are found inside the nucleus, not outside.

3.) Protons have a positive charge.

5 0

3 years ago

Read 2 more answers

Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m

Phantasy [73]

Answer:46.05 mm

Explanation:

Given

$Power=260\ hp\approx 260\times 746=193.96\ KW$

speed $N=550\ rev/min$

allowable shear stress $(\tau )_{max}=15\ kpsi\approx 103.421\ MPa$

Power is given by

$P=\frac{2\pi NT}{60}$

$193.96=\frac{2\pi 550\times T}{60}$

$T=3367.6\ N-m$

From Torsion Formula

$\frac{T}{J}=\frac{\tau }{r}-----1$

where J=Polar section modulus

T=Torque

$\tau$ =shear stress

For square cross section

$r=\frac{a}{2}$

where a=side of square

$J=\frac{a^4}{6}$

Substituting the values in equation 1

$\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}$

$a=0.04605\ m$

$a=46.05\ mm$

7 0

3 years ago

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0

3 years ago

A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock

rewona [7]

Answer:

31.25 meters or ~31 meters approximately

Explanation:

Let's see which of the 5 variables we are given since this is a constant acceleration problem.

$v_i \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \triangle x \\ a$

We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.

We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².

We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.

Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, $v_i=0 \frac{m}{s}$ .

We want to find the vertical displacement, the height of the bridge, so we can say that $\triangle x= \ ?$

We have 4 out of 5 variables:

$v_i,\ a, \ t, \ \triangle x$

Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):

$x_f=x_i+v_it+\frac{1}{2}at^2$

Subtract $x_i$ from both sides of the equation to get:

$\triangle x=v_it+\frac{1}{2}at^2$

Substitute in our known variables and solve for delta x.

$\triangle x=(0\frac{m}{s})(2.5s) + \frac{1}{2} (-10\frac{m}{s^2})(2.5s)^2$

0 m/s multiplied by 2.5 s is 0, so we have:

$\triangle x =\frac{1}{2} (-10)(2.5)^2$

Evaluate the exponent first and multiply the terms together.

$\triangle x =(-5)(6.25)$
$\triangle x =-31.25$

The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.

7 0

3 years ago

Read 2 more answers

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0

3 years ago