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antiseptic1488 [7]
4 years ago
10

if the DIY parachute is dropped freely from the rooftop of the building of a building and reached the ground 3s later.Upon reach

ing the ground,What is the (a)final velocity of DYI parachutes ?(b) Height of the rooftop?
Physics
1 answer:
Goshia [24]4 years ago
5 0

Answer:

It depends on what the parachute is made out of.

Explanation:

You might be interested in
How does energy relate to force?
ioda

Answer:work is force times distance

Explanation:to create a force u need energy and the greater the energy the greater the force is applied to an object.

4 0
3 years ago
A 400 hp engine in a 1,600 kg car applies maximum force for 2 seconds to accelerate the car onto the
babymother [125]

Answer:

I will assume that “maximum force” implies the constant application of power  P  = 400 hp (international) to accelerating the vehicle. The force will therefore vary with speed as the vehicle accelerates. I will also assume that all engine energy goes into accelerating the vehicle, rather than rotating elements like its wheels.

In this case the 400 hp (equivalent to 298,280 watts) is applied for time  t  = 2 seconds. Therefore the kinetic energy of the vehicle is increased by:

ΔKE=Pt=(298,280)(2)=596,560  joules.

The initial kinetic energy is:

KEinitial=12mv2

=(0.5)(1600)(82)=51,200  joules.

Therefore final kinetic energy is:

KEfinal=KEinitial+ΔKE

=51,200+596,560

=647,760  joules

Therefore final vehicle velocity can be found:

KEfinal=12mv2

v=2KEfinalm−−−−−−−−√

=(2)(647,760)1600−−−−−−−−−−−√

= 28.455 m/s

Explanation:

4 0
3 years ago
The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 84 MW at a cruising speed of 35.0 kn
frosja888 [35]

Answer:

F = 4669.201\,kN

Explanation:

Maximum speed is get at maximum power. Let assume that ship travels at constant speed, the expression for power is equal to:

\dot W = F\cdot v

Where F and v are the forward force and speed of ship measured in newtons and meters per second, respectively.

The forward force can be determined by clearing it in the expression described above:

F = \frac{\dot W}{v}

F = \frac{84\times 10^{3}\,kW}{(35\,knots)\cdot \left(\frac{0.514\,\frac{m}{s} }{1\,knot} \right)}

F = 4669.201\,kN

4 0
3 years ago
Read 2 more answers
A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
3 years ago
The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with th
aivan3 [116]

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

A=\frac{1}{12}m(3r^2+l^2)

Replacing the values we have,

A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)A=197.9kg.m^2

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

c=\frac{1}{2}mr^2

c=\frac{1}{2}(500)(0.5)^2

c=62.5kg.m^2

Finally we need to find the required angle between the fixed line a-a (I attached an image )

\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}

Replacing the values that we have,

\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}

\Phi = 2tan^{-1}(\sqrt{0.076634})

\Phi = 2tan^{-1}(0.2768)

\Phi = 2(15.47)

\Phi = 30.95\°

4 0
4 years ago
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