Answer:
1 * 10^-7 [J]
Explanation:
To solve this problem we must use dimensional analysis.
1 ergos [erg] is equal to 1 * 10^-7 Joules [J]
![1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]](https://tex.z-dn.net/?f=1%5Berg%5D%2A%5Cfrac%7B1%2A10%5E%7B-7%7D%20%7D%7B1%7D%2A%5B%5Cfrac%7BJ%7D%7Berg%7D%20%5D%20%5C%5C%3D%201%2A10%5E%7B-7%7D%5BJ%5D)
Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s
Answer:
<h3>30m</h3>
Explanation:
Velocity is the change of rate of displacement with respect to time.
velocity = displacement/time
Given
initial velocity = 15 m/s.
time taken =2 secs
Required
Displacement of the object
From the formula;
Displacement = Velocity * time
Displacement = 15 * 2
Displacement = 30m
<em>Hence the displacement of the object is 30m</em>
Give that,
The frequency range the dog can hear is 15Hz to 50,000Hz
The wavelength of sound in air at 20°C =?
Speed of sound is 344
The frequency corresponding to the lower cut-off point is the lowest frequency which his 15Hz
F=15Hz
The relationship between the wavelength, speed and frequency is given as
v=fλ
Then,
λ=v/f
λ=v/f
λ=344/15
λ=22.93m
You need to draw two lines from the head of the arrow: one parallel to the axis (which will be bent) and the other one passing through the center of the converging lens <span>(which will continue straight) </span>and see where they intersect.
If the arrow is at an infinite distance, the image will be point-like in F.
If the arrow is at a distance greater than 2F, the image will be upside down between F and 2F.
If the arrow is at a distance of 2F, the image will be upside down in 2F.
If the arrow is at a distance between F and 2F, the image will be upside down beyond 2F.
If the arrow is at a distance F, all rays proceed parallel to each other and no image is formed.
If the arrow is at a distance smaller than F, the image will be virtual.
Therefore, the correct matches are 1B, 2C, 3D, and 4A.
