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2 years ago

Three differences between orbital motion and axial motion of the earth.

Physics

1 answer:

NISA [10]2 years ago

7 0

Answer:

Orbital motion

1.is a gravitational motion.

2.they move in the same direction.

3.move are the same time.

Axial motion

1.is a rotary motion

2.move around it own axis

3. revolution takes one year

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A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha

Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

$PQ = \sqrt{4^{2}+3^{2}}=5 m$

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

$E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C$

Electric field at P due to the charge q is

$E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C$

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C$

3 0

2 years ago

Whipple is confused about the connection between the velocity and acceleration of the tennis ball. he decides to compare the vel

tamaranim1 [39]

The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.

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2 years ago

Describe a situation in which different units of measure could cause confusion.

forsale [732]

When an experiment happens in the USA and the test of the world read it conversions may go wrk v and think that the experiment is false

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2 years ago

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti

san4es73 [151]

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$

8 0

2 years ago

On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo

timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = $\frac{wf}{t}$ = $\frac{1.4 rad/s}{21 s} = 0.067 rad/s2$

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

3 0

2 years ago