Answer : The correct option is (d) 2.73 m
Explanation :
By the 2nd equation of motion,
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where,
s = distance or height = ?
u = initial velocity = 3.0 m/s
t = time = 0.5 s
a = acceleration due to gravity = ![9.8m/s^2](https://tex.z-dn.net/?f=9.8m%2Fs%5E2)
Now put all the given values in the above equation, we get:
![s=(3.0m/s)\times (0.5s)+\frac{1}{2}\times (9.8m/s^2)\times (0.5s)^2](https://tex.z-dn.net/?f=s%3D%283.0m%2Fs%29%5Ctimes%20%280.5s%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%289.8m%2Fs%5E2%29%5Ctimes%20%280.5s%29%5E2)
![s=2.73m](https://tex.z-dn.net/?f=s%3D2.73m)
Therefore, the correct option is (d) 2.73 m
Answer:
Mass and velocity.
Explanation:
Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:
![K=\frac{1}{2} mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
where
is the mass of the object, and
is its velocity at a given point in time.
So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:
- mass of the ball
- velocity of the ball before it hits the ground
With the knowledge of these two quantities the kinetic energy of the ball before touching the gound can be determined.
The loudness<span> of a sound is linked to the size of the vibration which produces it. A big vibration makes a louder sound. Scientists use the word '</span>amplitude<span>' for the size of waves. For waves on water, it is easy to measure the </span>amplitude<span>.</span>
Answer:
12.4 m/s²
Explanation:
L = length of the simple pendulum = 53 cm = 0.53 m
n = Number of full swing cycles = 99.0
t = Total time taken = 128 s
T = Time period of the pendulum
g = magnitude of gravitational acceleration on the planet
Time period of the pendulum is given as
![T = \frac{t}{n}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bt%7D%7Bn%7D)
![T = \frac{128}{99}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B128%7D%7B99%7D)
T = 1.3 sec
Time period of the pendulum is also given as
![T = 2\pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
![1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}](https://tex.z-dn.net/?f=1.3%20%3D%202%283.14%29%20%5Csqrt%7B%5Cfrac%7B0.53%7D%7Bg%7D%7D)
g = 12.4 m/s²