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Illusion [34]
2 years ago
8

If the mass of each ball is 12 kg and they experience a gravitational force of magnitude 2.00

Physics
1 answer:
VLD [36.1K]2 years ago
7 0

Answer:

{2 \times 10} - 7 = 12 {}^{2}  \div x \\ x = 2 \times 10 - 7 \div 12 {}^{2}

I'm just in jss2 but I read physics. this is what I think

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A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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What can be found in every skeletal muscle?
Talja [164]
C I’m not that sure though
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3 years ago
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Why does cold weather feel colder inside?
schepotkina [342]
One reason is that when you have been out in the cold, your hands feet and exposed features of your face will take time to recover as the blood circulation improves and supplied warm blood to capillaries. So the relatively warm room you enter will not immediately feel warm until the blood has regained its normal circulation. Other factors are that windows are cooled from the outside and condensation forms on the inside because of moisture in the air. For this condensation to evaporate requires heat, which will be extracted from the room and the air near the windows will be cooled. The cold air will descend and form a draught at floor level and this will tend to make the room cooler.
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3 years ago
A device used to measure the atmospheric pressure is the barometer. true false
Angelina_Jolie [31]
True, I just learned this a week ago. Is this for Chemistry?
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2 years ago
Study the roller coaster diagram. Match the labels to the correct location on the roller coaster.
IRISSAK [1]

Answer:

A- Greatest Kinetic Energy

B- Increasing Potential Energy

C- Increasing Kinetic Energy

D- Greatest Potential Energy

Explanation:

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2 years ago
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