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kicyunya [14]
3 years ago
7

One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chines

e in 1054. In 1968 it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angular speed in rad/s of the Crab nebula pulsar?
Physics
1 answer:
erica [24]3 years ago
7 0

Answer:

The angular speed of the Crab nebula pulsar is 190.3 rad/s.

Explanation:

Given that,

Time T= 33 ms = 0.033 s

The angular speed is equal to the 2π divided by time period.

We need to calculate the angular speed of the Crab nebula pulsar

Using formula of angular speed

\omega=\dfrac{2\pi}{T}

Where, T = time

\omega = angular speed

Put the value into the formula

\omega=\dfrac{2\pi}{0.033}

\omega=190.3\ rad/s

Hence, The angular speed of the Crab nebula pulsar is 190.3 rad/s.

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a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?
astra-53 [7]
We are given the following conditions:
v_{0} = 28m/s

t = 4s

v_{f} = 0m/s

Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.

So we look at our kinematics equation that contains the given variables.

v_f = v_0 + at

This contains our target variable and our initial conditions. Plug and chug, we get:

(0m/s) = (28m/s) + a(4s) -28m/s = (4s)*a a = -7m/s^2

Thus the answer is:

a = -7m/s^2

Hope this helps!


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3 years ago
What happened 1 billion years after the Big Bang?
maks197457 [2]

Explanation:

1 billion years after the big bang, the temperature is 20K and some stars and galaxies began to contract due to the gravitational contraction of the over densities of the previous universe. At about 10 billion years after the big bang, our earth and sun form

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3 years ago
A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction b
Leokris [45]

By Newton's second law,

• the net force acting vertically on the crate is 0, and

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0   ==>   <em>n</em> = <em>mg</em> = 1470 N

where <em>n</em> is the magnitude of the normal force; and

• the net force acting in the horizontal direction on the crate is also 0, with

∑ <em>F</em> = <em>f</em> - <em>b</em> = 0   ==>   <em>b</em> = <em>f</em> = <em>µn</em> = 0.645 (1470 N) = 948.15 N

where <em>b</em> is the magnitude of the braking force, <em>f</em> is (the maximum) static friction, and <em>µ</em> is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.

Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.

Let <em>a</em> be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for <em>a</em> :

<em>f</em> = <em>ma</em>   ==>   <em>a</em> = (948.15 N) / (150.0 kg) = 6.321 m/s²

With this acceleration, the truck will come to a stop after time <em>t</em> such that

0 = 50.0 km/h - (6.321 m/s²) <em>t</em>   ==>   <em>t</em> ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s

and this is the smallest stopping time possible.

8 0
3 years ago
A scientist who wants to study the affects of fertilizer on plants sets up an experiment. Plant A gets no fertilizer, Plant B ge
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Answer: Plant A

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zhuklara [117]

The most dangerous frequencies of electromagnetic energy are X-rays, gamma rays, ultraviolet light and microwaves. X-rays, gamma rays and UV light can damage living tissues, and microwaves can cook them. Hope this helps! =^-^=

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