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Anni [7]
2 years ago
7

NEED THIS ASAP

Physics
1 answer:
horrorfan [7]2 years ago
5 0

Answer:

Energy is essentially work done by an object or on object.

From,

W = Fd

It's directly proportional to mass.

from,

K. E = 1/2mv²

Energy is directly proportional to mass.

P. E = mgh

Energy is directly proportional to mass.

H = mc∆T

Energy is directly proportional to mass.

Thus increasing mass will increase the energy also imparted on another object since all the above eqns show that relationship.

And for 2 moving bodies

K.Ei = K.Ef(energy conservation)

m1u²1 + m2u²2 = m1v²1 + m2v²2

The relationship is the same that the greater mass the greater the impact.

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For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.
raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

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The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

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What is the net force on this object?
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Read 2 more answers
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
2 years ago
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