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3 years ago

15

A plane flies 125 km/hr at 25 degrees north of east with a wind speed of 36 km/hr at 6 degrees south of east. What is the result

ing velocity of the plane (in km/hr)?

1 answer:

nasty-shy [4]3 years ago

5 0

Answer:

V = 156.85 Km/h

Explanation:

Speed of plane = 125 Km/h

angle of plane= 25° N of E

Speed of wind = 36 Km/h

angle of plane = 6° S of W

Horizontal component of the velocity

V_x = 125 cos 25° + 36 cos 6°

V_x = 149 Km/h

Vertical component of the velocity

V_y = 125 sin 25° - 36 sin 6°

V_y = 49 Km/h

Resultant of Velocity

$V = \sqrt{V_x^2 + V_y^2}$

$V = \sqrt{149^2 + 49^2}$

V = 156.85 Km/h

the resulting velocity of the plane is equal to V = 156.85 Km/h

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Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion. Choose the planetesimal t

motikmotik

Answer:2

Explanation:

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4 years ago

Astronomers classify stars according to their (1 point) distance from Earth. color, size, and absolute brightness. age and paral

charle [14.2K]

Answer:

All of the above

Explanation:

Astronomers use all of those measures to classify stars. If you want to look more into classifying stars, check out the Hertzsprung-Russel Diagram. It covers how to identify red giants, main sequence, dwarf stars, ect. Distance from earth is typically measured in light years. The color of stars generally determines how hot they are. (Blue stars are the hottest) Also, the parallax method is used to measure stars that are closer to earth. This method relies heavily on geometry though.

Hope this helped!

8 0

3 years ago

A protostar's radius decreases by a factor of 100 and its surface temperature increases by a factor of two before it becomes a m

Sliva [168]

Answer:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

Explanation:

For this case we can use the formula of luminosity in terms of the radius and the temperature given by:

$L_i = K r^2 T^4$

Where L_i = initial luminosity, r= radius and T = temperature.

We know that we decrease the radius by a factor of 100 and the temperature increases by a factor of 2 so then the new luminosity would be:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

6 0

3 years ago

A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At w

mr_godi [17]

Answer:

θ₁ = 5.4°

θ₂ = 10.86°

Explanation:

The angle ca be found by using grating equation:

mλ = d Sinθ

where,

m = order of diffraction

λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m

d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line

θ = angle = ?

FOR m = 1:

(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₁ = 0.09425

θ₁ = Sin⁻¹(0.09425)

<u>θ₁ = 5.4°</u>

<u></u>

FOR m = 2:

(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₂ = 0.1885

θ₂ = Sin⁻¹(0.1885)

<u>θ₂ = 10.86°</u>

6 0

3 years ago

Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

2 years ago