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tatuchka [14]
2 years ago
12

How often do you rely on media for scientific information to make decisions in your life? (No weather answer please)

Physics
2 answers:
vodka [1.7K]2 years ago
6 0
The majority of the time


Nina [5.8K]2 years ago
4 0
Well, taking out the weather option ... I hardly ever do.
If I'm looking at a specific technical problem, then a scan on a source I'm not allowed to cite here, it seems, or some such may give me a hint as to whether it's going to be any help to me. If it is, i continue. If it's isn't, then i'll go away and decide whether I should pursue the problem or not (ie am I wasting my energy here ?)
I suspect, though, that I'm not representative of users.
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Answer:

Pretty sure its -4.8%

Explanation:

Others asked this question and got this as an answer.

5 0
2 years ago
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1. In the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7.
timofeeve [1]

Answer: n the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7. Calculate the proper life of the ...

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5 0
3 years ago
Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
klio [65]

Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

The centripetal acceleration of an object on a circular path is given by :

a=r\omega^2

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let r_1 is distance of Abbie from the merry-go-round and r_2 is distance of Zak's from the merry-go-round. Acceleration of Abbie is :

a_1=r_1\omega^2 ...... (1)

r_1=1\ m

Acceleration of Zak's is :

a_2=r_2\omega^2 .......(2)

r_2=2\ m

Dividing equation (1) and (2) we get :

\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}

So, the acceleration of Abbie is half of the Zak's.

7 0
3 years ago
A basketball has a mass of 1 kg and is traveling 15 m/s. How fast would a 5 kg bowling ball have to travel to have the same mome
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8 0
2 years ago
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A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N
Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s

Force on object is

F(t)=-mAw^{2}Cos(wt)

Substitute given values

So

F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N

So

Magnitude F(t)=26.6 N

Direction: -x

4 0
3 years ago
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