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3 years ago

6

Best budget freestyle drone?

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RUDIKE [14]3 years ago

4 0

Hx115 it's pretty fine looking

You might be interested in

Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

Tom [10]

Answer:

Explanation:

The pictures below shows the whole explanation for the problem

4 0

4 years ago

A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r

d1i1m1o1n [39]

Answer:

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

$a_c=\dfrac{F_c}{m}$

Thus, you can write the equations for each star:

$a_c_1=\dfrac{F_c_1}{m_1}$

$a_c_2=\dfrac{F_c_2}{m_2}$

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

$a_c_1\times m_1=a_c_2\times m_2$

Now you can clear $a_c_2$ :

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

6 0

4 years ago

An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is

devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= $2\pi\sqrt{\frac{L}{g}}$

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= $2\pi\sqrt{\frac{L}{9.81}}$

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = $\frac{\textup{AE}}{\textup{L}}$

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = $\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}$

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = $\sqrt{\frac{K}{m}}$

or

mass, m = $\sqrt{\frac{K}{\omega_n^2}}$

or

mass, m = $\sqrt{\frac{20.66\times10^8}{0.95^2}}$

mass, m = 22.90 × 10⁸ kg

4 0

3 years ago

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Det

suter [353]

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, $W_d_{(field)}$ = 18 kN/m³

maximum dry unit weight measured, $W_d_{(max)}$ = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

$RC = \frac{W_d_{(field)}}{W_d_{(max)}}$

substitute the given values;

$RC = \frac{18}{17} = 1.0588$

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

6 0

4 years ago

. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3

max2010maxim [7]

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

$Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)$

$Q_{losses} = 1.824\times 10^{6}\,kJ$

In order to keep the house warm, given heat must be equal to heat losses:

$Q_{H} = Q_{losses}$

Besides, the Coefficient of Performance for a Carnot heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

Where,

$T_{L}$ - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

$T_{H}$ - Temperature of the hot reservoir (House), measured in Kelvin.

Given that $T_{L} = 276.15\,K$ and $T_{H} = 295.15\,K$ , the Coefficient of Performance is:

$COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}$

$COP_{HP} = 15.534$

For a real heat machine, the Coefficient of Performance is determined by the following expression:

$COP_{HP} = \frac{Q_{H}}{W}$

Where:

$Q_{H}$ - Heat received by the house, measured in kilojoules.

$W$ - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

$W = \frac{Q_{H}}{COP_{HP}}$

$W = \frac{1.824\times 10^{6}\,kJ}{15.534}$

$W = 117419.853\,kJ$

The working time is calculated by dividing this result by input power. That is:

$\Delta t = \frac{W}{\dot W}$

$\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)$

$\Delta t = 3.624\,h$

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

8 0

4 years ago