Which form of energy is converted to electrical energy in a voltaic cell?.

What are the differences and relationships between speed, velocity, and acceleration

melamori03 [73]

Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.

3 0

3 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

7 0

2 years ago

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

expeople1 [14]

Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × $10^6$ kilometers

turn = 70°

travel an additional = 2.7 × $10^6$ kilometers

solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

Qx =2.7 × $10^6$ cos(70)

Qy= 2.7 × $10^6$ sin(70)

so that

Hx = Px + Qx ............1

Hx = 2.62 × $10^6$

and

Hy = Py + Qy ..........2

Hy = 2.53 × $10^6$

so Magnitude = $\sqrt{((2.62 \times 10^6 )^2+(2.53 \times 10^6)^2)}$

Magnitude = 3.64 ×

so direction will be

tan စ = Hy ÷ Hx ......................3

tan စ = $\frac{2.53}{2.62}$

tan စ = 0.9656

စ = 43.9°

3 0

3 years ago

You are skateboarding in your friend's driveway. You hit a rock and your

nekit [7.7K]

Answer:

the rock hit the board you go flying because you board stops but you have energy still going so there for you go flying

Explanation:

3 0

2 years ago

. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a long time?

galina1969 [7]

(a). The power of the candy bar is,

$P=440\text{ W}$

The time taken to play on 800 kJ energy of the candy bar is,

$t=\frac{E}{P}$

where E is the energy.

Substituting the known values,

$\begin{gathered} t=\frac{800\times10^3}{440} \\ t=1.818\times10^3\text{ s} \\ t=1818\text{ seconds} \end{gathered}$

As 1 minute is equal to 60 seconds,

Thus,

$\begin{gathered} t=1818\text{ seconds} \\ t=\frac{1818}{60} \\ t=30.3\text{ min} \end{gathered}$

Thus, the time taken to play tennis on the 800 kJ energy is 30.3 minutes.

(b). By doing the exercise, the process of digestion of food inside our body increases. Thus, the exercise does not helps us to burn the calories. But it helps us to diggest the heavy meal like candy bar easily.

The time taken to digest the canndy bar or to utilise its energy is large because it takes a lot of time to burn small amount of food and make it digest quickly.

8 0

1 year ago