Answer:
The specific gravity of the unkown liquid is 15.
Explanation:
Gauge pressure, at the bottom of the tank in this case, can be calculated from
where 
and 
are the height of the column of oil and the unkown liquid, respectively. Writing for 
, we have
Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.
Answer: 1175 J
Explanation:
Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."
Given
Spring constant, k = 102 N/m
Extension of the hose, x = 4.8 m
from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m
Work done =
W = 1/2 k [x(i)² - x(f)²]
Since x(f) = 0, then
W = 1/2 k x(i)²
W = 1/2 * 102 * 4.8²
W = 1/2 * 102 * 23.04
W = 1/2 * 2350.08
W = 1175.04
W = 1175 J
Therefore, the hose does a work of exactly 1175 J on the balloon
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.
Answer:
hi there
Explanation:
1 - III
2- 1
3-1
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Your rotational speed would still be the same. This is because all parts of the Ferris wheel rotate together. Your linear speed however would change. That is a function of radius. But the question is asking about rotational speed and that does not change in this situation
