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castortr0y [4]
1 year ago
7

Like all planets, the planet Venus orbits the Sun in periodic motion and simultaneously spins about its axis. Just as on Earth,

the time to make one complete orbit (i.e., the period of orbit) is what defines a year. And the time to make one complete revolution about its axis (i.e., the period of rotation) is what defines a day. The period of orbit for the Earth is 365.25 days and the period of rotation is 24 hours (1.00 day). But when these same values for Venus are expressed relative to Earth, it is found that Venus has a period of orbit of 225 days and a period of rotation of 243 days. So for Venus inhabitants, a day would last longer than a year! Determine the frequency of orbit and the frequency of rotation (in Hertz) on Venus. Ans: A marine weather station detects waves which are 9.28 meters long and 1.65 meters high and travel a distance of 50.0 meters in 21.8 seconds. Determine the speed and the frequency of these waves. Ans:
Physics
1 answer:
Liono4ka [1.6K]1 year ago
7 0

Answer:

a) F = 5.14 10⁻⁸ Hz,  f = 4.76 10-8 Hz,  b)   v = 2.29 m / s,   f = 42.5 Hz

Explanation:

a)This problem has two parts.

For the calculations relative to the planet Venus, we use that the period and the frequency are related

            f = 1 / T

frequency of the orbit around the Sun

   

Let's reduce the period to the SI system

           T = 225 days (24h / 1days) (3600 s / 1h) = 1.94 10⁷ s

           F = 1 / 1.94 10⁷

           F = 5.14 10⁻⁸ Hz

rotation frequency

            T = 243 d = 2.1 107 s

             f = 1 / T

             f = 1 / 2.1 107

            f = 4.76 10-8 Hz

b) give the data of some marine waves

the speed of the wave can be found with kinematics

            v = x / t

            v = 50.0 / 21.8

            v = 2.29 m / s

If the wavelength is L = 9.28m

this distance is the distance between two consecutive ridges or valleys

             λ / 2 = L

             λ = 2L

             λ = 2 9.28

             λ = 18.56 m

the speed of the wave is

             v = λ f

             f = v /λ

             f = 2.29 / 18.56

             f = 42.5 Hz

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The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle
serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

6 0
2 years ago
A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


5 0
2 years ago
It is easier to climb up a slanted slope than a vertical slope
V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0
2 years ago
Read 2 more answers
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At lef
Sladkaya [172]

Question:

Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. In which direction should an arrow at position 1 point?

left

right

up

down

Answer:

The correct answer is

Left

Explanation:

Magnetic circuits describe the path of a magnetic flux. In the same way electricity follows a complete closed circuit, the path of a magnetic flux is also a complete and closed circuit which leaves from the N pole, migrates through the air  and reenters the magnet through the S pole through which it passes back into the magnet to come to the N pole again.

As such the magnetic field lines emanate from the N pole which is on he right to the S pole which is on the left. Hence the arrow should point in the left direction.

3 0
2 years ago
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Capillary action in trees can transport water from the roots to the tree's branches. The capillaries (Xyelem) in a certain tree
lesantik [10]

Answer:

h=14.2857\,m

Explanation:

Given:

radius of capillary, r=10^{-6}\,m

angle of contact, \theta=0^{\circ}

density of water, \rho=1000\,kg.m^{-3}

surface tension of water, T=0.07 \,N.m^{-1}

height, h = ?

We have the equation for the height of meniscus as:

h=\frac{2T.cos\, \theta}{\rho.g.r}

h=\frac{2\times 0.07\times cos\,0^{\circ}}{1000\times 9.8\times 10^{-6}}

h=14.2857\,m

No, the capillary action alone cannot be the mechanism of water transportation to the top of the trees. Transpiration also creates a suction pressure in the xylem complementary to the ascent of sap and cohesion of water being the other causes of movement of water up in the plants.

6 0
2 years ago
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