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gladu [14]
3 years ago
8

I'LL MARK U BRAINLIST!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Roman55 [17]3 years ago
8 0
The answer is B.
hope this helps!

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Which phrase best describes the time period in which the current body of scientific knowledge was developed? A. Just over the la
LiRa [457]

Answer:

The correct option is;

B. Throughout all of history

Explanation:

A body of knowledge comprises of a given collection of concepts, activities and terms that represents a professional discipline that is outlined by the associated scholarly society, therefore, scientific knowledge, being comprised of very diverse field of knowledge, with different approaches and  changing understanding is taking as a process of learning about the universe data collection, experimenting analysis and communication with a theme of systems, equilibrium, models and constant change which makes the basis of the body of scientific knowledge to be throughout all of history.

3 0
3 years ago
A 6.2 kg object moving in the +x direction at 5.3 m/s collides head-on with an 7.8 kg object moving in the −x direction at 2.5 m
Crank

Answer:

10ms kg 88.2

Explanation:

Toook test

8 0
3 years ago
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
Which of the following is a heterogeneous mixture?
Reil [10]
That would be Italian dressing because it does not mix completely therefore it is the definition of a heterogeneous mixture
8 0
3 years ago
A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
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