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2 years ago

What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months

Physics

1 answer:

iren2701 [21]2 years ago

7 0

Answer:

93312000 Joules

Explanation:

Energy can be explained as the capacity for doing work. And can be expressed as

E= pt....................eqn(1)

Where

E= Energy in Joules

P= power in Watt= 2.00W

t= time in seconds

Time= 18 months, which can be converted to "seconds" as;

18 months = [ 18 monthsx 30 days x 24 hours x 60 min x 60 sec] = 46656000

Then

time= 46656000 seconds

Power=2 watt

If we substitute into equation (1) we have

Energy = 2 watt × 46656000 seconds

Energy= 93312000 Joules

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PLEASE HELP!! ITS URGENT!!!

Natasha2012 [34]

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

$(m_{1}*v_{1}) - F*t = (m_{1}*v_{2})$

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

$(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]$

Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.

6 0

2 years ago

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)

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The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

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Given that

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Determine , Vertical component = v sin Θ

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