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Dimas [21]
3 years ago
11

It is dangerous to stand near the railway leak when train passing by,why?​

Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

Because a person may be pulled in the direction of the moving train. Thereby causing accident

Explanation:

According to Daniel Bernoulli's theorem, he was widely known as a Mathematician. He stated that due to the higher velocity of a moving train, there is higher kinetic energy in terms of volume around it, while the air pressure between the person and the train becomes lower.

As a result, a person near a moving train may be pulled in the direction of the moving train. Thereby causing accidents that may lead to death.

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Well i think it is false caus ebunch of products these day have metals incorparded in the product
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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b
daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a)  Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 1.5 m      

     Substituting

                      s = ut + 0.5 at²

                      1.5 = 0 x t + 0.5 x 9.81 xt²

                      t = 0.553 s

      Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 200 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 0.553 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 200 x 0.553 + 0.5 x 0 x 0.553²

                      s = 110.6 m

      The bullet travels horizontally 110.6 m

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3 years ago
If the refractive index of a medium is 1.3,
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Explanation:

see the above attachment to solve the question and get the answer.

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3 years ago
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A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
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Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)
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3 years ago
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