What is a scientific paradigm
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Alpha particles \ beta particles / gamma radiation Put these different types of radiation in order from MOST to LEAST penetrating.
Q2. Zara travelled in a train moving at an average speed of 120km/h and covered a distance of 40km towards East. Calculate the time taken by the train to cover this distance.
What is the density of water?gram/mL(fill in the number)
A rock at the edge of a 200-m high cliff has more potential energy than an equal sized rock at the edge of a 600-m high cliff. True or False
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B.False
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Answer:
A.True
A ship is able to float on water because It is less dense than the water.
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A geostationary orbit, also known as a geosynchronous equatorial orbit, is a specific type of circular orbit around the Earth. In this orbit, a satellite's orbital period matches the Earth's rotation period, which is approximately 24 hours. This causes the satellite to appear stationary relative to a fixed point on the Earth's surface, which is why it's commonly used for communication, weather, and broadcasting satellites.
To achieve this, the satellite needs to be at a specific height above the Earth's surface. The altitude at which a satellite will have a period of one day (24 hours) is called the geostationary altitude or the geostationary orbit height.
The calculation to determine this altitude involves using the formula for the orbital period of a satellite:
Where:
- \( T \) is the orbital period (24 hours in this case)
- \( G \) is the gravitational constant
- \( M \) is the mass of the Earth
- \( r \) is the radius of the satellite's orbit (distance from the center of the Earth to the satellite)
Solving for \( r \) with the known values of \( T \), \( G \), and \( M \), you'll find that the satellite needs to be approximately 35,786 kilometers (22,236 miles) above the Earth's surface to have an orbital period of one day.
This altitude allows the satellite to complete one orbit around the Earth in the same amount of time it takes for the Earth to complete one rotation on its axis, effectively staying in the same position relative to a specific point on the Earth's equator.
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It decreases 9.8 meters per second each second, until it reaches
its highest point and starts falling. Then its speed increases
9.8 meters per second each second, until it hits the ground.
(2) a ball falling freely toward the surface of
Earth
(3) a car moving with a constant speed along a
straight, level road
(4) a projectile at the highest point in its trajectory
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3) a car moving with a constant speed along a
straight, level road.
Hopefully this helps and good luck
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Can you calculate the force on the electron ?
Sure ! It's pushing against a force of 180 Newtons per Coulomb,
and you know the charge on it.
Do you know the distance it's moved ?
Sure ! That's the 2.5 cm (=0.025 meter) between the plates.
Look at this:
Work = (force) x (distance)
= (180 N/C) x (1.602 x 10⁻¹⁹ C) x (0.025 m)
I get 7.209 x 10⁻¹⁹ N-m. (joules) Try it and see if you agree.