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3 years ago

If the mass of an object i if the mass of an object is 44 kg in its velocity is 10 m/s East how much kinetic energy

Physics

1 answer:

vampirchik [111]3 years ago

8 0

K=0.5 mu×u
K=2200J no matter the direction

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If a transformer has 5000 turns on the primary circuit, 3000 turns on

nikitadnepr [17]

The answer is C , you’re welcome

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2 years ago

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

3 years ago

An engine has an imput of 100 j and gives out 27 j of useful kenetic energy what is the efficiency of the engine

Lunna [17]

Answer:

Based on what we have learned so far, in what ways have the experiences of Asian-Canadians stayed the same over time? Why do you think these similarities exist? Please give specific examples to support your ideas.

Explanation:

4 0

2 years ago

What is the weight on mass

Gre4nikov [31]

Answer:

6.39×10^23 kg is the weight on mass

3 0

1 year ago

The quantity of charge passing through a surface of area 1.82 cm2 varies with time as q = q1 t 3 + q2 t + q3 , where q1 = 5.2 C/

Brrunno [24]

Answer:

Current through the surface at t = 1.1 s is 21.37 A.

Explanation:

The charge is passing through a surface of area varies with time as :

$q=q_1t^3+q_2t+q_3$

Here,

$q_1=5.2\ C/s^3\\\\q_2=2.5\ C/s\\\\q_3=6.5\ C$

t is in seconds

$q=5.2t^3+2.5t+6.5$

The rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dq}{dt}\\\\I=\dfrac{d(5.2t^3+2.5t+6.5)}{dt}\\\\I=15.6t^2+2.5$

At t = 1.1 s, Current,

$I=15.6(1.1)^2+2.5\\\\I=21.37\ A$

So, the instantaneous current through the surface at t = 1.1 s is 21.37 A.

3 0

2 years ago