Answer:
Water is not able to be used as a thermometer liquid because of its higher freezing point and lower boiling point than the other liquids in general. If water is used in a thermometer, it will start phase variation at 0∘C and 100∘C. This will not help in measuring temperature, beyond this range.
Explanation:
plzzzzzzz Mark my answer in brainlist
Answer:
ac = 3.92 m/s²
Explanation:
In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,
Frictional Force = Centripetal Force
where,
Frictional Force = μ(Normal Force) = μ(weight) = μmg
Centripetal Force = (m)(ac)
Therefore,
μmg = (m)(ac)
ac = μg
where,
ac = magnitude of centripetal acceleration of car = ?
μ = coefficient of friction of tires (kinetic) = 0.4
g = 9.8 m/s²
Therefore,
ac = (0.4)(9.8 m/s²)
<u>ac = 3.92 m/s²</u>
To solve this we assume
that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 =284.15 x 2.50 / 303.15
<span>V2 = 2.34 L</span>
Answer:
Option A.
A fan is turned from high speed to low speed.
Explanation:
It is important to note that air is also a fluid.
In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.
Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.
This makes option A the correct option
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
