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15 days ago

In an effort to protect a rhino, volunteers are following its steps with air monitoring and ground cameras. The rhino starts on

day 1 from ground camera A and walks for 1.5 km west along a straight line as seen from the air. The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B. In this location the rhino may find water and food, therefore the animal stays for the night. On the second day, the rhino moves 2.5 km directly south as recorded from the air by the volunteers. In these two days of tracking how far and in which direction does the rhino travel from camera A?

Physics

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A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Softa [3030]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Reservoir pressure = 10 atm

$T_1$ = Reservoir temperature = 300 K

$P_2$ = Exit pressure = 1 atm

$T_2$ = Exit temperature

$R_s$ = Specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

Assuming isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

Flow density at exit measures 2.2721 kg/m³

4 0

3 months ago

The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports

kicyunya [3294]

To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02

An alternative method:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Thus, they must fly at 82.02 mph.

5 0

1 month ago

Read 2 more answers

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = $\sqrt{2I/ce_{0} }$

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

1 month ago

In the past, salmon would swim more than 1130 km (700 mi) to spawn at the headwaters of the Salmon River in central Idaho. The t

Keith_Richards [3271]

Answer:

$E_t_o_t_a_l=7.603MJ$

Explanation:

The overall energy expenditure of the salmon, which corresponds to its swimming upstream effort, $W$ , is linked to its specific mechanical power. $Mechanical \ power$ calculated per unit mass can be derived from the following equation:

$\frac{p}{m}=\frac{1}{m}|\frac{dW}{dt}|=2W/kg\\\frac{1}{2}|\frac{W}{22\times 24 \times 60\times 60}|=2\\\\W=7.603MJ$

As a result, the total energy utilized during the 22-day journey is 7.603 MJ

6 0

2 months ago