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4 years ago

True or false? A protostellar cloud spins faster as it contracts

Physics

1 answer:

Aleks04 [339]4 years ago

3 0

Answer:

No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)

Explanation:

Once the cloud is so dense that the heat which is being produced in its center cannot easily escape, pressure rapidly rises, and catches up with the weight, or whatever external force is causing the cloud to collapse, and the cloud becomes stable, as a protostellar cloud.

The protostellar cloud will become more dense over thousands of years. This stage of decreasing size is known as a contraction, rather than a collapse. In the contraction stage the cloud has become much slower, and because weight and pressure are more or less in balance. In the first stage of formation, the decrease of size is very rapid, and compressive forces completely overwhelm the pressure of the gas, and we say that the cloud is collapsing.

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Ball A is allowed to fall and strike ball B. Assume that all of ball A's energy is transferred to ball B at point I, and that th

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Therefore the Kinetic Energy of Ball B equals the Total Energy of Ball A minus the Potential Energy of Ball A

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A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning

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The tiger would not be able to produce glucose causing it to stop running

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3 years ago

Two charged objects have a repulsive force of 0.040 N. If the distance separating the objects is doubled, then what is the new f

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Two charged objects have a repulsive force of 0.080 N. If the charge of both of the objects is doubled, then what is the new force? Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of both objects is doubled, then the force will become four times greater.

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

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Magnitude of acceleration is -43200 ft/s²

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3 years ago

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