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3 years ago

List the forms of renewable energy that are curently in use

1 answer:

4 0

Here are the ones that I know about
and can think of just now:

-- wind
-- solar
-- nuclear
-- tidal
-- hydro
-- geothermal
-- biomass

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The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45

sesenic [268]

<h2>Speed required is 12.05 m/s</h2>

Explanation:

Let the velocity of puma be a.

Consider the vertical motion of puma

We have equation of motion v² = u² + 2as

Initial velocity, u = asin45

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Displacement, s = 3.7 m

Substituting

v² = u² + 2as

0² = (asin45)² + 2 x -9.81 x 3.7

a² = 145.19

a = 12.05 m/s

Speed required is 12.05 m/s

4 0

3 years ago

Please help ASAP for the fill the gap sentences. The brackets are for the energy stores..... thank you in advance.

lesya692 [45]

The answer is c or h

4 0

4 years ago

Calculate the period of wave whose frequency is 50Hz

miss Akunina [59]

Explanation:

time period= 1/frequency.

= 1/50 = 0.02 second.

hope this helps you.

5 0

3 years ago

Read 2 more answers

A 0.42 kg shuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/s at constant accelerat

MariettaO [177]

Answer

given,

mass of the shuffleboard disk = 0.42 kg

speed of the cue is increased to = 4.2 m/s

acceleration takes over 2 m then acceleration is zero.

the disk additionally slide to 12 m

final speed of disk = 0 m/s

a) increase in thermal energy

$\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)$

$\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)$

$\Delta E_t = 3.704\ J$

b) $\Delta E_t = F_f.d$

F_f is the frictional force

$3.704= 12.F$

$F_f = 0.308\ N$

increase in thermal energy for entire movement of 14 m

$\Delta E_t = 0.308\times 14$

$\Delta E_t =4.312 \ J$

c) Work done on the disk by the cue

$W = \Delta KE + \Delta E_{t}$

$W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d$

$W = 3.704+ 0.308 \times 2$

W = 3.704 + 0.616

W = 4.32 J

3 0

3 years ago

Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ

mash [69]

The energy stored in a capacitor is given by
$U= \frac{1}{2} CV^2$
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.

Let's calculate the energy of the first capacitor:
$U_1 = \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J$

And now the energy of the second capacitor:
$U_2 = \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J$

So, the total energy stored in the two capacitors is
$U=U_1 +U_2 = 1.8 \cdot 10^{-3}J$

3 0

3 years ago

Read 2 more answers