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ludmilkaskok [199]
3 years ago
10

Un hombre de pie puede ejercer la misma fuerza con sus piernas tanto en la tierra como en la luna. Sabemos que la masa del hombr

e es la misma en la Tierra y en la Luna. También sabemos que F=m x a (masa x aceleración, corresponde a la 2° Ley de Newton), es verdad tanto en la Tierra y la Luna. ¿Será el hombre capaz de saltar más alto en la Luna que en la Tierra? ¿Por qué o por qué no?
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

The height jumped by the person on the moon is 6 times the height jumped by the person on earth.  

Explanation:

As we know that the acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth.

So, it is false.

Let the mass of man is m and the gravity on moon is g' = g/6.

Let the height jumped on earth is h and the height jumped on moon is h'.

So,

m x g' x h' = m x g x h

g/6 x h' = g x h

h' = 6 h  

So, the height jumped by the person is 6 times the height jumped by the person on earth.

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A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t
Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s

and the positive sign means the initial direction was rightward.

6 0
3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

         fr = 52.1 N

Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

        N - W₁ -W₂ = 0

        N = W₁ + W₂

as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

          fr₁ = 52.1 N

          fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

          F = 156.3 N

7 0
2 years ago
Question 18 according to newton's second law of motion, the force acting on an object varies directly with the object's accelera
leonid [27]
Solve for "x"
X=force
18/6=x/9
cross multiply 
162=6x
x=27
Hope this helps
3 0
3 years ago
Describe the point at which the kinetic energy of the pendulum is the highest
Arlecino [84]
The highest point<span> of the </span>pendulums<span> swing is when the potential energy is at its </span>highest<span> and the </span>kinetic energy<span> is at its lowest.</span>
5 0
3 years ago
A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc
tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we have

d_i = 2.3 m = 230 cm

d_o = 25 cm

now plug in all values into the formula

-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}

f = 28 cm

now for power of lens

P = \frac{1}{f}

P = \frac{1}{0.28} = 3.5 D

so the power in dioptre is

P = 3.5 D

5 0
3 years ago
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