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Vlada [557]
3 years ago
7

How can you say that joule and Newton are derived units​

Physics
1 answer:
maxonik [38]3 years ago
3 0

Answer:

F = M a    where M is acceleration and a is acceleration

a = x / s^2 = distance / time squared

The Newton is derived because mass, distance, and time are all fundamental units One would have to look at the fundamental requirements for these definitions, but they can all be repeated in a laboratory.

So the Newton is determined from these fundamental units and since the Joule equals Newton * Distance it is also derived from the fundamental units.

If one has the three fundamental units then one can derive the Joule and Newton.

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Answer:

A and C. UvU

Explanation:

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Help asap!! The resultant force on the box is
rodikova [14]

The resultant force on the box is C) 10 Newtons to the left.

Explanation:

The figure of the problem is attached.

We see that there are two forces acting on the box along the horizontal direction:

- A force of 10 N acting to the right

- A force of 20 N acting to the left

By taking "to the right" as positive direction, we can rewrite the two forces as follows:

F_1 = +10 N\\F_2 = -20 N

Where the second force is negative since it acts to the left.

Since the two forces are in line, we can find their resultant simply by calculating their algebraic sum. By doing so, we find:

F=F_1 + F_2 = +10 + (-20) = -10 N

And the negative sign tells us that the direction is to the left.

Therefore, the correct answer is

C) 10 Newtons to the left.

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brainly.com/question/12978926

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3 years ago
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A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
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