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3 years ago

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an air craft is travelling due north with velocity of 100m/s astrong wind blow from the west with in velocity of 25m/s the reshs

tant velocity

1 answer:

OleMash [197]3 years ago

8 0

Explanation:

The aircraft is traveling north at 100 m/s.

The wind blows from the west (towards the east) at 25 m/s.

The two vectors form a right triangle. The magnitude of the resultant velocity can be found with Pythagorean theorem:

v² = vx² + vy²

v² = (25 m/s)² + (100 m/s)²

v = 103 m/s

The direction can be found with trigonometry:

θ = atan(vy / vx)

θ = atan(100 / 25)

θ = 76.0°

The resultant velocity is 103 m/s at 76.0° north of east.

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If you do 1500 J of work hoisting a 20 kg bale of hay , to what height did you lift it

strojnjashka [21]

Explanation:

W = PE

W = mgh

1500 J = (20 kg) (9.8 m/s²) h

h = 7.65 m

Round as needed.

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3 years ago

Pls help meh! im in 5 grd<br><br> what is used to see white light?<br><br> pls respond asap!

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3 years ago

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of

Sauron [17]

Answer:

The value is $t = 689.029 \ hours$

Explanation:

From the question we are told that

The molar mass of hydrazine is $Z = 32 g/mol = \frac{32}{1000} = 0.032 \ kg/mol$

The initial temperature is $T_i = -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K$

The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

The specific heat capacity is $c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)$

The power available is $P = 300 \ W$

The mass of the fuel is $m = 1640 \ kg$

Generally the number of moles of hydrazine present is

$n = \frac{m}{Z}$

=> $n = \frac{1640}{= 0.032}$

=> $n = 51250 \ mol$

Generally the quantity of heat energy needed is mathematically represented as

$Q = n * c_h * (T_f -T_i)$

=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

=> $Q = 7.441516913 * 10^{8} \ J$

Generally the time taken is mathematically represented as

$t = \frac{Q}{P}$

=> $t = \frac{7.441516913 * 10^{8} }{300}$

=> t = 2480505.6377 s

Converting to hours

$t = \frac{2480505.6377}{3600}$

=> $t = 689.029 \ hours$

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

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3 years ago

If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th

Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

σ = Stefan-Boltzmann constant

ε = Emissivity

A = Area

T = Absolute Temperature

Therefore,

A = E/σεT⁴

So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

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3 years ago