Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
Answer: A light bulb can be all of the following except option C (a consumer product if it is used to light the office of the board of directors.)
Explanation:
Products are classified as being BUSINESS or CONSUMER products according to the buyer's intended use of the product.
-Consumer products: these are sold goods that are used for personal, family, or household use. The intention of the buyer is for the products to satisfy his personal needs and desires. Example of some of the consumer products include: toothpaste, eatables and clothes.
Business products: products that are not for personal use but for the manufacturing of other goods are called business products.
Therefore a bulb is not serving as a personal use when used to light the office of the board of directors rather it's serving as a business product .
1. friction between water molecules
2. the wave spreads out onto a larger and larger area, so per unit area, the energy of the wave goes down
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol