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mamaluj [8]
3 years ago
12

A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is

not negligible. Identify the forces on the falcon.
Physics
1 answer:
abruzzese [7]3 years ago
7 0

Answer:

As the falcon begins to fall towards the ground, then there will be two forces acting on it, 1) weight of the falcon acting downward and 2) the drag force acting on it due to it's motion through the air.

Explanation:

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A rocket consists of a shuttle and a fuel tank. The rocket flies through the air with a momentum of 180000kgm/s at a velocity of
babunello [35]

Answer:

Answer is 1000kg

Hope it help

6 0
2 years ago
Show that the Mass spring system executes simple harmonic motion(SHM)?​
murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0
3 years ago
A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W
Neko [114]

Answer:

W = 0.49 N

τ = 0.4851 Nm

Force

Explanation:

The weight force can be found as:

W = mg

W = (0.05 kg)(9.8 m/s²)

<u>W = 0.49 N</u>

The torque about the pivot can be found as:

τ = W*d

where,

τ = torque

d = distance between weight and pivot = 99 cm = 0.99 m

Therefore,

τ = (0.49 N)(0.99 m)

<u>τ = 0.4851 Nm</u>

The pivot exerts a  <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.

6 0
3 years ago
What are some forces that come in pairs?
Nikolay [14]

Answer:

ACTION REACTION FORCES

Explanation:

When there is an action frce there will be a reaction force

3 0
3 years ago
A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0
3 years ago
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