Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

we can express the velocity by the derivative of the displacement,
Hence

at equilibrium, the velocity becomes

Hence if we substitute values we arrive at

Answer:
40N
Explanation:
Since both weights are connected to one string, you can say that the tensions above each are equal to each other.
If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.
If I were to put my process into steps:
- Write out the sum of Forces for both masses
- Set them equal to each other to find normal force (because this is the only unknown)
- Calculate and compare the two tensions to see if they are equal
*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.
Since the y axis stayed consistent, we can assume it did not move at all.
(So your answer would be A)
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 )
(a) Zero
The maximum efficiency (Carnot efficiency) of a heat engine is given by

where
is the low-temperature reservoir
is the high-temperature reservoir
For the heat engine in the problem, we have:


Therefore, the maximum efficiency is

(b) Zero
The efficiency of a heat engine can also be rewritten as

where
W is the work performed by the engine
is the heat absorbed from the high-temperature reservoir
In this problem, we know

Therefore, since the term
cannot be equal to infinity, the numerator of the fraction must be zero as well, which means
W = 0
So the engine cannot perform any work.