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3 years ago

What feature of a battery affects how often it needs to be replaced?

Physics

1 answer:

Gnoma [55]3 years ago

4 0

Answer:

c.lifespan or battery lifespan

Explanation:

"Battery Lifespan"affects how often it needs to be replaced

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A 3-kg object is attached to a spring and moving in simple harmonic motion. Its angular frequency is 20 rads/sec. When the mass-

Trava [24]

Answer:19.5 J

Explanation:

Given

mass of block=3 kg

angular frequency=20 rad/sec

spring constant $k=\omega _n^2m=1200 N/m$

we know total energy remain conserved

$E_T at x=0.1 m$

$E_T=E_P+E_K$

Where $E_K$ =kinetic energy

$E_P$ =potential Energy

$E_P=\frac{1}{2}kx^2$

$E_P=600\times 0.01=6 J$

$E_K=\frac{1}{2}mv^2$

$E_K=\frac{1}{2}\times 3\times 3^2=13.5 J$

$E_T=13.5+6=19.5 J$

When mass reaches amplitude its velocity becomes zero

there is only potential energy which is equal to Total energy

$E_T=E_P=19.5 J$

7 0

3 years ago

Which electromagnetic wave enables us to see objects

Darina [25.2K]

Visible light because it enables us to see light and nklngdjofjjovjojiofhidhzfkbfjbjdbjbfueuishfihehfnih

6 0

3 years ago

Please help

V125BC [204]

Answer:

her arm when she is throwing it

Explanation:

because that is the action

7 0

3 years ago

Lift a notebook a few inches off a table or other surface. As it moves up, it has kinetic energy. Which force opposed the motion

Greeley [361]

If im correct the answer is gravity
and wind resistance if you lift it fast

~~~hope this helps~~~
~~davatar~~

3 0

3 years ago

Read 2 more answers

A plastic block of mass 0.20.2kg is initially at rest on a horizontal frictionless surface. A bullet of mass 1212 g is fired wit

Juli2301 [7.4K]

Answer:

94.29619%

Explanation:

$m_1$ = Mass of bullet= 0.012 kg

$m_2$ = Mass of block = 0.2 kg

$u_1$ = Initial Velocity of bullet= 550 m/s

$u_2$ = Initial Velocity of block = 0 m/s

$v_1$ = Final Velocity of bullet = 20 m/s

$v_2$ = Final Velocity of block

As the linear momentum of the system is conserved

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{0.012\times 550+0.2\times 0-0.012\times 20}{0.2}\\\Rightarrow v_2=31.8\ m/s$

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(m_1u_1^2-m_1v_1^2-m_2v_2^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(0.012\times 550^2-0.012\times 20^2-0.2\times 31.8^2)\\\Rightarrow \Delta K=1711.476\ J$

Percentage change is given by

$\dfrac{\Delta K}{\dfrac{1}{2}m_1u_1^2}\times 100=\dfrac{1711.476}{\dfrac{1}{2}0.012\times 550^2}\times 100=94.29619\ \%$

The percent of the initial kinetic energy of the bullet block system is lost during the collision is 94.29619%

4 0

3 years ago