Assuming constant acceleration <em>a</em>, the object has undergoes an acceleration of
<em>a</em> = (50 m/s - 100 m/s) / (25 s) = -2 m/s²
Then the net force has a magnitude <em>F</em> such that, by Newton's second law,
<em>F</em> = (75.0 kg) <em>a</em>
<em>F</em> = (75.0 kg) (-2 m/s²)
<em>F</em> = -150 N
meaning the object is acted upon by a net force of 150 N in the direction opposite the initial direction in which the object is moving.
Answer:
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Answer:
Q = 590,940 J
Explanation:
Given:
Specific heat (c) = 1.75 J/(g⋅°C)
Mass(m) = 2.01 kg = 2,010
Change in temperature (ΔT) = 191 - 23 = 168°C
Find:
Heat required (Q)
Computation:
Q = mcΔT
Q = (2,010)(1.75)(168)
Q = 590,940 J
Q = 590.94 kJ
Answer:
Amplitude increases with decreasing velocity.
Explanation:
At the same time, an increase in attention takes place
Answer:
The rate of flow of water is 71.28 kg/s
Solution:
As per the question:
Diameter, d = 18.0 cm
Diameter, d' = 9.0 cm
Pressure in larger pipe, P = 
Pressure in the smaller pipe, P' = 
Now,
To calculate the rate of flow of water:
We know that:
Av = A'v'
where
A = Cross sectional area of larger pipe
A' = Cross sectional area of larger pipe
v = velocity of water in larger pipe
v' = velocity of water in larger pipe
Thus
v' = 4v
Now,
By using Bernoulli's eqn:
where
h = h'
Now, the rate of flow is given by:
