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Len [333]
3 years ago
13

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Physics
1 answer:
Lisa [10]3 years ago
5 0

Answer:

....

Explanation:

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Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.
Vikentia [17]
The Newton’s law Nikolas would use to come up with this idea is the <span>Third law that states:

</span><span>When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
</span>
So, in this case, let's name the first Body A which is the skateboard and the second body B which is <span>the compressed carbon dioxide in a fire extinguisher. Then, as shown in the figure below, according to the Third law:

</span>FA = -FB<span>

</span>

8 0
3 years ago
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Identify the temperature that is equivalent to 95°F. Use this formula to convert the temperature
Stells [14]
The answer is 35 degrees Celsius. Hope I helped :) Please vote brainliest. 
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The work accomplished (produced) by a machine is called Work _____
Nonamiya [84]
I think it’s output because output work is work done by a machine
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Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of
Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San  Francisco i.e 400 mi

considering ,

Alan's speed v_A=50mph

Beth's speed v_B=60mph

->For Alan:

The time required t_A= d/v_A= 400/50 => 8h

-> For beth:

The time required t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0
3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

\alpha=\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

The angular velocity can be written as

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

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

T'=T+8.09\cdot 10^{-6} =0.01400809 s

Therefore, the change in angular velocity after 1 year is

\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s

So, the angular acceleration of the pulsar is

\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y

b)

To solve this part, we can use the following equation of motion:

\omega'=\omega + \alpha t

where

\omega' is the final angular velocity

\omega is the initial angular velocity

\alpha is the angular acceleration

t is the time

For the pulsar in this problem:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s is the initial angular velocity

\omega'=0, since we want to find the time t after which the pulsar stops rotating

\alpha = -0.259 rad/s/y is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y

which means that the period of the pulsar increases by

\Delta T=8.09\cdot 10^{-6} s

For every year:

\Delta t=1 y

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

T=T_0+\frac{\Delta T}{\Delta t}\Delta t

where T_0 is the original period and

\Delta t=869 y

is the time that has passed; solving for T0,

T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s

6 0
3 years ago
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