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3 years ago

A box is being pushed from the left with a force of 5 N. No force is being applied from the right.

Physics

1 answer:

Sergio039 [100]3 years ago

6 0

Answer:

The box will move to the right with increasing speed

Explanation:

According to Newton's second law:

$F=ma$ (1)

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the box in this problem, there is only one force acting on it: a force of 5 N from the left (so, to the right). This means that according to (1) the box will experience an acceleration: the direction of the acceleration is the same as the direction of the net force (so, to the right), therefore the box will move to the right and its speed will continue increasing due to fact that it is being accelerated.

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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.

Lift increases proportionally to the square of the speed.

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).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

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4 years ago

Which option defines green computing?

Papessa [141]

Study and practice of environmental sustainable computing or It. such as monitors, printers, storage devices, and networking and communication systems-efficiently and effectively with minimal impact on the environment.

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3 years ago

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Which property of a substance is not affected by physical changes?

Leokris [45]

The answer is A. Reactivity

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3 years ago

A motorcycle moving 18.8 m/s has<br> 57800 J of KE. What is its mass?

Brums [2.3K]

Answer:

m = 327.07 kg

Explanation:

Given that,

Kinetic energy of a motorcycle, E = 57800 J

Velocity of the motorcycle, v = 18.8 m/s

We need to find the mass of the motorcycle. The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

m is mass

$m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg$

So, the mass of the motorcycle is 327.07 kg.

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3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

wolverine [178]

Answer:

final velocity = 0.08585m/s

Explanation:

We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.

The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after

total linear momentum before = linear momentum of first train car + linear momentum of second train car

We know that linear momentum = mv

where,

m = mass

v = velocity

thus,

total linear momentum before = m₁v₁ + m₂v₂

m₁ = mass of first train car = 135,000kg

v₁ = velocity of first train car = 0.305m/s

m₂ = mass of first second car = 100,000kg

v₂ = velocity of second train car = −0.210m/s

Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.

total linear momentum before = m₁v₁ + m₂v₂

= 135,000x0.305 + 100,000x(−0.210)

= 135,000x0.305 - 100,000x0.210

= 20,175 kgm/s

Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).

Therefore, total linear momentum after = mv

m = m₁ + m₂ = 135,000 + 100,000 = 235,000

total linear momentum before = total linear momentum after

235,000v = 20,175

v = $\frac{20,175}{235,000}$

= 0.08585m/s

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3 years ago

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