Answer:
(a) decrease
Explanation:
Viscosity is the resistance which occur to flow of the fluid.
More the inter molecular forces between particles of the liquid, more the viscosity of liquid.
<u>Effect of temperature on viscosity:-</u>
Viscosity decreases with the increase in the temperature as forces among the particles decrease on increasing temperature. The kinetic energy of the particles of the liquid increases causing to move in more random motions and thus weaker inter molecular forces and this offer less resistance to the flow.
<u>Hence, viscosity of the liquids decrease with the increasing temperature.</u>
Answer:
(a) ω = 1.57 rad/s
(b) ac = 4.92 m/s²
(c) μs = 0.5
Explanation:
(a)
The angular speed of the merry go-round can be found as follows:
ω = 2πf
where,
ω = angular speed = ?
f = frequency = 0.25 rev/s
Therefore,
ω = (2π)(0.25 rev/s)
<u>ω = 1.57 rad/s
</u>
(b)
The centripetal acceleration can be found as:
ac = v²/R
but,
v = Rω
Therefore,
ac = (Rω)²/R
ac = Rω²
therefore,
ac = (2 m)(1.57 rad/s)²
<u>ac = 4.92 m/s²
</u>
(c)
In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:
Centripetal Force = Frictional Force
m*ac = μs*R = μs*W
m*ac = μs*mg
ac = μs*g
μs = ac/g
μs = (4.92 m/s²)/(9.8 m/s²)
<u>μs = 0.5</u>
Answer:
by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2
Answer:
Acceleration = 4.8 m/s²
Explanation:
Given:
Change in velocity = 19 m/s
Change in time = 4 s
Find:
Acceleration
Computation:
Acceleration = Change in velocity / Change in time
Acceleration = 19/4
Acceleration = 4.8 m/s²
Positive acceleration
Answer:
Explanation:
Formula
W = I * E
Givens
W = 150
E = 120
I = ?
Solution
150 = I * 120 Divide by 120
150/120 = I
5/4 = I
I = 1.25
Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)
On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.
However I would go with the other answerer's post and multiply both values by √2