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Xelga [282]
2 years ago
6

What do waves transfer

Physics
1 answer:
vichka [17]2 years ago
3 0

Answer:

disturbance I think not sure thou:)

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Explanation:

6 0
3 years ago
The slow coning pattern of the spin axis of the Earth's precession will move the end of this axis in a complete circle in a peri
GREYUIT [131]

Answer:

26000 years

Explanation:

Precession describes the angular motion of the Earth's body. Since the attitude of telescopes relative to the Earth's body can be controlled with high accuracy, and telescopes can measure the direction of incoming light also with high accuracy, the motion of Earth is under permanent high precision monitoring. Thus the basic numerical descriptor of precission, an angular rate of 5029.0966 seconds of arc per Julian century, traditionally denoted p (for precession) is a measured value from observed coordinate changes of thousands of stars over, say, two centuries. The understanding of this value in terms of forces acting on an oblate Earth from the Moon is well understood so that an extrapolation back and forth over a few full cycles contains little uncertainties. Of course, you can find details on the coordinate transformations mentioned above (the direct observational effect of precession) on the net. I was surprised to see that the Wikipedia article on precession covers the astronomical aspect very poorly. You thus better look for other sources.

3 0
4 years ago
How is a cold air mass similar to a mountain range as to the effect on a warm air mass?
Fittoniya [83]
Different types of fronts produce different patterns of weather. When a cold, dense air mass pushes warmer air, it produces a cold front. A cold front can turn a heat wave into normal summer weather or turn cold winter air into very cold weather. Cold fronts often produce tall cumulonimbus clouds.
8 0
3 years ago
I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?
Slav-nsk [51]

Answer:

The height from which the egg is dropped is <u>68.54 m</u>.

Explanation:

Given:

Initial velocity of egg is, u=0\ m/s(Dropped means initial velocity is 0)

Time taken is, t=3.74\ s

Acceleration experienced by egg is due to gravity, a=g=9.8\ m/s^2

The height from which the egg is dropped is, d=?

Now, we use Newton's equation of motion that relates the distance, initial velocity, time and acceleration.

So, we have the following equation of motion:

d=ut+\frac{1}{2}at^2

Plug in all the given values and solve for 'd'. This gives,

d=0+\frac{1}{2}\times 9.8\times (3.74)^2\\\\d=\frac{1\times 9.8\times 13.9876}{2}\\\\d=\frac{137.078}{2}\\\\d=68.54\ m

Therefore, the height from which the egg is dropped is 68.54 m.

8 0
3 years ago
A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
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