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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

You might be interested in

Two trains leave a train station traveling in the same direction. They leave at the same time. Train A moves at 41.1 mph and Tra

hammer [34]

The time taken for Train A to be 39.1 miles ahead of Train B is 2.7 hours

<h3>How to determine the relative speed</h3>

Speed of Train A = 41.1 mph
Speed of Train B = 26.5 mph
Relative speed =?

Relative speed = 41.1 - 26.5

Relative speed = 14.6 mph

<h3>How to determine the time to be 39.1 miles apart</h3>

Relative speed = 14.6 mph
Distance = 39.1 miles
Time =?

Relative speed = distance / time

14.6 = 39.1 / time

Cross multiply

14.6 × time = 39.1

Divide both side by 14.6

Time = 39.1 / 14.6

Time = 2.7 hours

Thus, it will take train A 2.7 hours to be 39.1 miles ahead of Train B

Learn more about speed:

brainly.com/question/14334174

#SPJ1

8 0

1 year ago

What is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes

enot [183]

Answer:

a= v -u\ t

650- 230\ 30

= 420\30

=14 m\s²

8 0

3 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

Which of the following statements about energy is correct? A.Energy is a vector quantity. B.Energy is the force used to move an

TiliK225 [7]

C. Energy is required to set an object in motion.

4 0

3 years ago

Read 2 more answers

As it pulls itself up to a branch, a chimpanzee accelerates upward at 2.4 m/s2 at the instant it exerts a 260-N force downward o

Helen [10]

Answer:

$F=208.83N$

Explanation:

From the question we are told that:

Acceleration $a=2.4m/s^2$

Force of Branch $F=260N$

Generally the Newton's equation second law for Force is mathematically given by

$ma=F-mg$

$m=\frac{260}{2.4+9.8}$

$m=21.31kg$

Therefore

$F=mg$

$F=(21.31)(9.8)$

$F=208.83N$

4 0

3 years ago