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lisov135 [29]
3 years ago
11

Can someone plz help me ASAP

Physics
2 answers:
d1i1m1o1n [39]3 years ago
8 0

Answer:

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Soloha48 [4]3 years ago
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What does activation energy has to do with a chemical reaction.
asambeis [7]

Explanation:

Activation energy and reaction rate

The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. ... The released energy helps other fuel molecules get over the energy barrier as well, leading to a chain reaction.

6 0
3 years ago
Which of the following is equivalent to 140 centiliters?
tekilochka [14]
D, 0.140 liters! Hang on a sec and I'll show you a trick I use.
7 0
3 years ago
Read 2 more answers
The third one pls help
Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 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 jet of water of cross section area and velocity v impinges normally on a stationary flat plate the mass per unit volume of wat
Sergio039 [100]
<h2>F = kAρv²</h2>

Explained in the attachment !

<h3>Hope it helps you!!</h3>

4 0
2 years ago
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