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3 years ago

How much do the all stars make on sytycd?

Physics

1 answer:

tigry1 [53]3 years ago

8 0

5089 I believe is the correct answer

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1. high-mass main-sequence stars have a more distant and wider than low-mass stars.

jenyasd209 [6]

High mass are largest , hotest , and brightest sequence

8 0

3 years ago

A disk, initially rotating at 120 rpm, is slowed down with a constant acceleration of magnitude 4 rad s2. If the disk has diamet

andrey2020 [161]

Answer:

Explanation:

given,

ω₁ = 120 rpm

1 rpm = $\dfrac{2\pi }{60} rad/s$

$\omega_1 =120\times \dfrac{2\pi }{60}$ rad/s

= 12.56 rad/s

α = - 4 rad/s²

diameter of disk = 20 cm

final angular velocity = 0

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

t = $\dfrac{0-(12.56)}{-4}$

t = 3.14 s.

2) $\theta = \omega_i t +\dfrac{1}{2} \alpha t^2$

= $12.56\times 3.14 +\dfrac{1}{2}\times (-4)\times 3.14^2$

= 19. 72 radians

3) total angular distance rotated

x = θ R

x = 19.72 × 0.1 = 1.97 m

x = 2 m

3 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

4 years ago

A wave travels with a speed of 78m/s in air and it has a frequency of 42hz.What is the wavelength of this wave?

stiks02 [169]

Answer:

1.9 m

Explanation:

78 m/s / (42 cycles/s) = 1.8571... m/cycle

5 0

3 years ago

A compact car has a maximum acceleration of 2.0 m/s2 when it carries only the driver and has a total mass of 1100 kg . you may w

nadya68 [22]

According to Newton`s law. Force exerted by car,

$F = m a = 1100 kg \times 2 m/s^2 = 2200 \ N$

After adding an additional 400 kg of mass, the force will be same therefore the acceleration

$F = 2200 \ N = (1100 \ kg + 400 \ kg) a \\\\ a = \frac{2200 \ N}{1500 \ kg} = 1.47 \ m/s^2$

Thus, the acceleration after adding the masses is 1.47 \ m/s^2.

4 0

3 years ago