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atroni [7]
2 years ago
10

Which statement describes Kepler’s third law of orbital motion?

Physics
1 answer:
garik1379 [7]2 years ago
8 0

Answer:

The square of orbital period is proportional to the cube of the semi-major axis.

Explanation:

I just took the quick check

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Use 3rd law to explain how a fish swims through water. Identify the force pairs and draw a diagram.
vovikov84 [41]
Well as the fish swims he pushes the water behind him which in return push him forward
5 0
2 years ago
Your car burns gasoline as you drive up a large mountain. What energy transformation is the car performing?
Sonja [21]

Your car is performing a transformation of energy of:

Chemical energy to Mechanical energy

The chemical is the gasoline which is then converted to fire as the car runs thus creating the movement of the car which is mechanical energy.

7 0
2 years ago
Read 2 more answers
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height
stira [4]

Answer:

9.5 kg m^2/s

Explanation:

The angular momentum of an object is given by:

L=mvr

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

L=(0.35)(9.0)(3.0)=9.5 kg m^2/s

4 0
3 years ago
Calcite (CaCO_3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field paralle
katrin2010 [14]

Answer:

(a) 42.28°

(b) 37.08°

Explanation:

From the principle of refraction of light, when light wave travels from one medium to another medium, we have:

\frac{n_{b} }{n_{a} } = sinθ_{a}/sinθ_{b}

In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.

For critical angle θ_{a} = θ_{c}, θ_{b} = 90°; n_{b} = 1

(a) n_{a} = 1.4864

\frac{1 }{1.4864 } = sinθ_{c}/sin90°

0.6728 = sinθ_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584

\frac{1 }{1.6584} = sinθ_{c}/sin90°

0.60299 = sinθ[tex]_{c}

θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°

7 0
3 years ago
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