All categories

3 years ago

Which of the following scientists won a noble prize of pioneering work in the study of the evolution of stars?

Physics

2 answers:

zhenek [66]3 years ago

7 0

Subrahmanyan chandrasekhar

Svet_ta [14]3 years ago

5 0

Subramanyan Chandrasekhar and William Alfred Fowler won a noble prize of pioneering work in the study of the evolution of stars.

The prize was divided equally between both scientists in 1983.

You might be interested in

A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?

ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0

3 years ago

Read 2 more answers

Some children are underfed and often have no place to sleep. According to

jeka94

Answer:

Physiological needs

Explanation:

3 0

3 years ago

What is the mass of something that weighs 300 N on earth

jeka57 [31]

Answer:

30.5810 kg

Explanation:

6 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$