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Tema [17]
4 years ago
5

Definition of magnitude

Physics
2 answers:
tiny-mole [99]4 years ago
6 0

Answer:

The great size or extent of something.

Explanation:

Hope this helps!

Harman [31]4 years ago
5 0

Answer:

<h2>the great size or extent of something!</h2>

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The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an
SashulF [63]

Answer:

 L = 41.09 Kg m2 / s      The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

        L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

      p = m v

Let's replace

      L = m r x v

The given function is

      x = 6.00 i ^ + 4.15 t j ^

We look for speed

     v = dx / dt

     v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

     L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]

      L = m 6 4.15 k ^

The other products give zero

Let's calculate

      L = 1.65 6 4.15 k ^

      L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0
3 years ago
What is energy converted to if it is not used to do work?
Romashka [77]

Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

3 0
2 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
The ratio of output power to input power, in percent, is called?
Harrizon [31]
That's efficiency. There's no law that it must be stated in percent.
4 0
3 years ago
Read 2 more answers
In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel
slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

v=60\times \dfrac{1609.34}{3600}=26.822\ m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s

or

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s

The time taken is 2.73414 seconds

The potential energy is given by

U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J

The change in potential energy is 467622.66798 J

5 0
3 years ago
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