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:
331.75 V
Explanation:
Given:
Number of turns of the coil, N = 40 turns
Area, A = 0.06 m²
Magnetic Field, B = 0.4 T
Frequency, f = 55 Hz
Maximum induce emf, E₀ = NABω
but ω = 2πf
Maximum induce emf, E₀ = NAB(2πf₀)
Maximum induce emf, E₀ = 2πNABf₀
Where;
N is number of turns of the coil
A is area
B is magnetic field
ω is the angular velocity
f is the frequency
E₀ = 2 × π × 40 × 0.06 × 0.4 × 55
E₀ = 342.81 V
The maximum induced emf is 331.75 V
Researchers have estimated that ______ to ______ genes reside within the chromosomes and
influence all genetic characteristics.
a. 10,000; 15,000
b. 20,000; 25,000
c. 50,000; 70,000
d. 100,000; 200,000 b. 20,000; 25,000 Answer is b. 20,000; 25,000
Answer: Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.
Explanation: In the 1600s, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
