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lutik1710 [3]
3 years ago
14

1. The horizontal and vertical components of a projectile's velocity are

Physics
1 answer:
Anni [7]3 years ago
5 0

The horizontal and vertical components of a projectile's velocity are independent of each other.

Answer: Option C

<u>Explanation:</u>

The path of a projectile is determined by two components of motion. They are termed as horizontal and the vertical components. Since both components velocity are perpendicular to each other, so it can stated that they are independent of each other.

Even it can seen that when the horizontal components of velocity is constant, then there will be change in the vertical components of velocity leading to free fall projectile path.

And in the absence of gravity, there will be change in the horizontal components of velocity with zero vertical component of velocity. Thus, the horizontal and the vertical components of a projectile’s velocity are seemed to be independent of each other.

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Answer:

The value  is   a =  2.7183

Explanation:

From the question we are told that

   The  relationship between the number of whooping cranes and the number of decades is ln N =  2.85 + 0.039t

  The  exponential relationship is  N  =  na^{kt}

Now from the given equation we have that

       N  =   e^{2.85 + 0.039t}

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So comparing this equation obtained an the given  exponential relationship we have that

      n = 17.29

       k  =  0.039

        a =  e =  2.7183

        a =  2.7183

 

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A lemming running 2.22m/s runs off a horizontal cliff. It lands in the water 7.45m from the base of the cliff. How much time was
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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
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Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

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Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

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From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

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Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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