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frozen [14]
3 years ago
12

I need help on putting this diagram in order.

Physics
1 answer:
morpeh [17]3 years ago
3 0
In what type of order are you supposed to put it in?
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A reconnaissance plane flies 605 km away from
kolezko [41]

Answer:

                      v_{avg}  = 355 m/s  

Explanation:

Distance = 605 km

Initial speed = v_{i} = 284 m/s

Final velocity = v_{f} = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

                        2as = v_{f}^{2}+v_{i}^{2}

Then using first equation of motion, we find time

                        v_{f} = v_{i}+at

Then using the formula of average velocity, we find average velocity

                         v_{avg}=\frac{total-distance}{total-time}

Second method is very simple

                                  v_{avg}=\frac{v_{f}+v_{i} }{2}

                                   v_{avg}=\frac{426+284}{2}

                                   v_{avg}  = 355 m/s      

8 0
3 years ago
During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio
Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where v = 0 m/s is the final angular velocity of the wheel when it stops, \omega_0 = 95rad/s is the initial angular velocity of the wheel, \alpha = -29.23 rad/s^2 is the deceleration of the wheel, and \Delta \theta is the angle swept in rad, which we care looking for:

0 - 95^2 = 2*29.23\Delta \theta

9025 = 58.46 \Delta \theta

\Delta \theta = 9025 / 58.46 = 154.375 rad

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

8 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
3 years ago
If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
creativ13 [48]

Answer:

-112.876J

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

16KNO_3(s) + 24C(s) + S_8(s)    \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)

Let us define P - V work as;

w_{pv} = - P_{external}  \triangle Volume

where  \triangle (Volume) = (V_{final} - V_{initial})

External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2

K_2S is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  CO_2  and  N_2 into grams using their respective molar masses.

0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

7 0
3 years ago
A table is moved using 60 N of force.
dem82 [27]

Answer:

15m

Explanation:

W=f×s

s =  \frac{w}{f} \\ s =  \frac{900j}{60n} \\ s = 15m

5 0
3 years ago
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