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3 years ago

Which unit is equivalent to j/s

1 answer:

4 0

To be honest, there's no sure way to answer that, because you haven't defined your terms and we can't be sure of what j or s might be.

Tell you what I'll do:. I'll assume definitions for j and s, and then I'll answer the question that I invented.

Assume that j stands for Joule, the unit of energy. And assume that s stands for 'second', the unit of time.

Then j/s is the rate of transferring energy or doing work.

Its unit is the Watt, equivalent to 1 Joule per second.

In your system of notation, it would be 'w' .

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Question 2

ra1l [238]

Answer:

0.729 meters

Explanation:

d=(V0)t+(1/2)at^2

d=height of shelf

a=9.8 m/s^2 (acceleration due to gravity)

V0=0 m/s (initial downward velocity)

t=0.38586 seconds (1.173 meters/3.04 m/s)

4 0

3 years ago

Beyond what point must an object be squeezed for it to become a black hole

Naddik [55]

Answer:

Condition

$\frac{m}{R} \geq \frac{c^{2} }{2*g}$

Explanation:

Basically black hole is an object from which light rays can not escape it means to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface = $-\frac{G*M*m}{R}$

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

$\frac{m*v^{2} }{2}$ = $-\frac{G*M*m}{R}$

⇒ $\frac{m}{R} =\frac{c^{2} }{2*G}$

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

4 0

3 years ago

Examine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.

serg [7]

Answer:

B) Absorbed energy results in the change in potential energy.

Explanation:

i got it wrong on usatestpreo

3 0

3 years ago

Read 2 more answers

Mass is a fundamental quantity

Oliga [24]

Explanation:

okk hope you like then comment

5 0

3 years ago

A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

$\vec{F}=q\vec{v}\times \vec{B}$

Thus, the magnitude of the force that the proton experiences is given by:

$F=qvBsin\theta$

The magnetic field is perpendicular to the proton's velocity, therefore, we have $\theta=90^\circ$ . Replacing the given values, we obtain:

$F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N$

3 0

3 years ago