That is a really good question, cheese is stretchy when it is hot is because when you heat it up, it liquefies which makes it stretch. it doesn't stretch when it is cold because it is a solid and solids usually do not stretch.
The correct option is C.
J.J Thompson demonstrated that Dalton's model of the atomic theory was wrong by showing that atoms are made up of sub particles, this is contrary to the report that Dalton gave about atoms. In his experiment, Dalton reported that atoms are indivisible, that is, they can not be broken down into smaller particles. J.J Thompson on the other was able to show through his cathode ray experiment that atoms are made up of smaller particles called electrons.
Answer:
Direction remains the same but velocity changes.
Explanation:
This tell us about the direction and magnitude of the acceleration acting on the cannonball throughout its duration of flight that its direction remains the same but its magnitude of the acceleration is continuously changing. The cannonball moves in the direction in which the cannon was fired while the velocity is highest after the fire but decreases when goes higher and when it comes back to the ground so its velocity increases against so we can say that both positive and negative acceleration occurs. Positive acceleration means increase in the magnitude of velocity whereas negative acceleration means decrease in velocity.
This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as ![T^{2} \alpha R^{3}](https://tex.z-dn.net/?f=T%5E%7B2%7D%20%5Calpha%20R%5E%7B3%7D)
⇒
Here K is the proportionality constant.
If
and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-
![\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%20%5E%7B2%7D%20%7D%7BT_%7B2%7D%20%5E%7B2%7D%20%7D%20%3D%5Cfrac%7BR_%7B1%7D%20%5E%7B3%7D%20%7D%7BR_%7B2%7D%20%5E%7B2%7D%20%7D)
⇒ ![R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }](https://tex.z-dn.net/?f=R_%7B1%7D%20%5E%7B3%7D%20%3DR_%7B2%7D%20%5E%7B3%7D%20%2A%5Cfrac%7BT_%7B1%7D%20%5E%7B2%7D%20%7D%7BT_%7B2%7D%20%5E%7B2%7D%20%7D)
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as
and
respectively.
Let the distance from sun and orbital period of earth is denoted as
and
respectively.
we are given that![T_{1} =29.46 years](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D29.46%20years)
we know that
1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-
![R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }](https://tex.z-dn.net/?f=R_%7B1%7D%20%5E%7B3%7D%20%3DR_%7B2%7D%20%5E%7B3%7D%20%2A%5Cfrac%7BT_%7B1%7D%20%5E%7B2%7D%20%7D%7BT_%7B2%7D%20%5E%7B2%7D%20%7D)
=![[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU](https://tex.z-dn.net/?f=%5B1%20%5D%5E%7B3%7D%20%2A%5Cfrac%7B%5B29.46%5D%5E%7B2%7D%20%7D%7B%5B1%5D%5E%7B2%7D%20%7D%20AU)
![=867.8916 AU^{3}](https://tex.z-dn.net/?f=%3D867.8916%20AU%5E%7B3%7D)
⇒![R_{1} =\sqrt[3]{867.8916}](https://tex.z-dn.net/?f=R_%7B1%7D%20%3D%5Csqrt%5B3%5D%7B867.8916%7D)
=9.5386 AU [ans]
The time of flight of the launched particle will be 2v₀ sinθ / g.
<h3>What is kinematics?</h3>
The study of motion without considering the mass and the cause of the motion.
A projectile is launched from level ground with an initial speed v₀ at an angle θ with the horizontal. If air resistance is negligible.
Then the time of flight of the launched particle will be given as,
Determine how long it requires the projectile to reach its highest point in order to calculate the flight duration. Just twice as long as the maximum altitude is the travel time.
![\rm v_y = v_{o_y} + a_y t](https://tex.z-dn.net/?f=%5Crm%20v_y%20%3D%20v_%7Bo_y%7D%20%2B%20a_y%20t)
At maximum height,
.
The time to reach maximum height is given as
![\rm t_{1/2} = - \dfrac{v_{o_y} }{ a_y}](https://tex.z-dn.net/?f=%5Crm%20t_%7B1%2F2%7D%20%3D%20-%20%5Cdfrac%7Bv_%7Bo_y%7D%20%7D%7B%20a_y%7D)
The time of flight is given as,
![\rm t = 2 \times t_{1/2} = - \dfrac{2v_{o_y}}{ a_y}](https://tex.z-dn.net/?f=%5Crm%20t%20%3D%202%20%5Ctimes%20t_%7B1%2F2%7D%20%3D%20-%20%5Cdfrac%7B2v_%7Bo_y%7D%7D%7B%20a_y%7D)
Substituting in
, then we have
The time of flight of the launched particle will be 2v₀ sinθ / g.
More about the kinematics link is given below.
brainly.com/question/7590442
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