I need help on putting this diagram in order.

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;

P sun = I sun-earth A sun-earth

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

P sun = 2.863 10²³ × 1360

P 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;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

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

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 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 $1.00$ $atm$ , 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