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garik1379 [7]
3 years ago
13

The Earth rotates toward the east. True False

Physics
2 answers:
elena-s [515]3 years ago
5 0

Answer:

True.East

Explanation:

nikdorinn [45]3 years ago
4 0

Answer:

true

Explanation:

earth rotates eastward, in pro grade motion. as viewed from the north pole star polaris, earth turns counter clockwise. the north pole, also known as the geographic north pole or terrestrial north pole, is the point in the northern hemisphere where earth's axis of rotation meets its surface.

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What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?
iren2701 [21]

Answer:

  • 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

V(r) = \frac{4}{3} \ \pi \ r^3

If we got an uncertainty \Delta r the formula for the uncertainty of V is:

\Delta V(r) = \sqrt{  (\frac{dV}{dr} \Delta r)^2  }

We can calculate this uncertainty, first we obtain the derivative:

\frac{dV}{dr}  = 3 * \frac{4}{3} \ \pi \ r^2

\frac{dV}{dr}  = 4 \ \pi \ r^2

And using it in the formula:

\Delta V(r) = \sqrt{  (4 \ \pi \ r^2\Delta r)^2  }

\Delta V(r) = \sqrt{  4^2 \ \pi^2 \ r^4 \Delta r^2  }

\Delta V(r) =  4 \  \pi \ r^2 \Delta r

The relative uncertainty is:

\frac{\Delta V(r)}{V(r)}

\frac{ 4 \  \pi \ r^2 \Delta r  }{ \frac{4}{3} \ \pi \ r^3}

\frac{ 3  \Delta r  }{  r}

Using the values for the problem:

\frac{ 3 * 0.09 m  }{  5.66 m} = 0.0477

This is, a percent uncertainty of 4.77 %

4 0
2 years ago
What are non-contact forces please include example in your answer
Alina [70]

Answer:

Gravitational force.  Magnetic force.  Electrostatics.  Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0
3 years ago
Read 2 more answers
What property or properties of a wave determine it's speed?
mash [69]
The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.
7 0
3 years ago
I have a special thermos that has a vacuum chamber between my hot soup and its outside wall. Because my soup is isolated from it
bagirrra123 [75]

Answer:

C. My soup will cool down because of radiation.

Explanation:

Since, there is vacuum between hot soup and its outside wall, then heat can not flow through conduction and convection.

The heat then only flows through radiation.

Therefore, the soup will not cool down because of convection or conduction by because of radiation.

7 0
3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
2 years ago
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