A continental polar air mass forms in ___

A solid ball and a hollow ball, each with a mass of 1.00 kg and radius of 0.100 m start from rest and roll down a ramp of length

nordsb [41]

Answer:

The solid ball and hollow ball both will reach the bottom with the same speed.

Explanation:

The speed of the solid and hollow balls is independent of the mass and the radius. A solid and hollow ball experience same speed on a given incline.

The speed can be calculated as

v = √(10/7)gh

where g is gravitational acceleration and h is the height

sinθ = h/L

h = L*sinθ

h = 3*sin(35)

h = 1.72 m

v = √(10/7)*9.8*1.72

v = 4.91 m/s

Both balls will reach the bottom at the speed of 4.91 m/s.

8 0

3 years ago

How far can a car traveling at a rate of 40 kilometer per hour travelbin 2 1/2 hours

CaHeK987 [17]

90 kilometers because you need to multiply 40 by 2 and then you get 80 and finally you add 10 and get 90 kilometers

4 0

3 years ago

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago

The earth and the moon exert forces on each other which forces is greater? explain

Helen [10]

Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.

An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

acentr=(4 pi2 r)/T2

Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.

This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.

8 0

3 years ago

If the period of a wave is 0.05 s, then what is its frequency?

yKpoI14uk [10]

Frequency = 1/time period = 1/0.05 = 20s^-1.

8 0

3 years ago

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A continental polar air mass forms in ____