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NISA [10]
3 years ago
6

Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima

que puede alcanzar el carrito?
Physics
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

5.241\ \text{m/s}

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = 9.81\ \text{m/s}^2

h = Altura = 1.4\ \text{m}

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}

La velocidad máxima que puede alcanzar el coche es 5.241\ \text{m/s}.

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Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t
miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0
3 years ago
What is one advantage of doing a field experiment instead of a laboratory experiment
Flauer [41]

It mimics the real world accurately

Explanation:

Experiments conducted in the field clearly presents the real world at it is to the scientist. Hardly can any part be controlled precisely and this gives a near to perfect scenario.

  • In the laboratory, for example, an organism is isolated from its environment and might not fully display its natural instinct and physiological capabilities.
  • Most laboratory set up are driven towards a model instead of real life settings.
  • The laboratory is more controlled and less varied and might truly represent the real world. It will only portray a part of the real world and series of further tests might have to be carried out to have a better model.

Learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

5 0
3 years ago
Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?
Mademuasel [1]
The first opiton is the answer A)<span>Rahul’s weight
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8 0
3 years ago
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Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The c
m_a_m_a [10]

Answer:

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

6 0
3 years ago
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An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon
erastova [34]

The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ_{1} = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ_{2} = Adjacent side 2 / Hypotenuse

cos θ_{2} = 4.98 / 8.5 = 0.59

θ_{2} = cos^{-1} ( 0.59 )

θ_{2} = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

To know more about trigonometric ratios

brainly.com/question/1201366

#SPJ1

3 0
1 year ago
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