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1 year ago

how much energy would microraptor gui have to expend to fly with a speed of 10 m/sm/s for 1.0 minute?

Physics

1 answer:

Ray Of Light [21]1 year ago

3 0

Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.

Learn more about Microraptor gui here brainly.com/question/1200755

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What could be the possible answer to the question ? thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately 0.377·M·g

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ 0.377·M·g

Learn more about equilibrium of forces here:

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2 years ago

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A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0

ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

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3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

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2 years ago

A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin

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Answer: The total momentum before the docking maneuver is $mV_{1}+MV_{2}$ and after the docking maneuver is $(m+M) U$

Explanation:

Linear momentum $p$ (generally just called momentum) is defined as mass in motion and is given by the following equation:

$p=m.v$

Where $m$ is the mass of the object and $v$ its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision $p_{i}$ will be the same as the total momentum of these same two objects after the collision $p_{f}$ ".

$p_{i}=p_{f}$

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

$p_{i}=mV_{1}+MV_{2}$

Where:

$m$ is the mass of the vehicle

$V_{1}$ is the velocity of th vehicle

$M$ is the mass of the space station

$V_{2}$ is the velocity of the space station

And the final momentum is:

$p_{f}=(m+M)U$

Where:

$U$ is the velocity of the vehicle and space station docked

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The proportion of carbon-14 in an organism is useful in figuring out the age of that organism after it dies because...?

nata0808 [166]

I believe the correct answer from the choices listed above is option D. The proportion of carbon-14 in an organism is useful in figuring out the age of that organism after it dies because the proportion of carbon-14 slowly decreases after the death of the organism. Hope this answers the question.

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