Answer:
Difference in height = 7.5 cm
Explanation:
We are given;.
Height of ethyl alcohol;h2 = 20 cm = 0.2 m
Density of glycerin: ρ1 = 1260 kg/m³
Density of ethyl alcohol; ρ2 = 790 kg/m³
To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.
Thus;
P1 = P2
Formula for pressure is; P = ρgh
Thus;
ρ1 × g × h1 = ρ2 × g × h2
g will cancel out to give;
ρ1 × h1 = ρ2× h2
Making h1 the subject, we have;
h1 = (ρ2× h2)/ρ1
h1 = (790 × 0.2)/1260
h1 = 0.125 m
Difference in height will be;
Δh = h2 - h1
Δh = 0.2 - 0.125
Δh = 0.075 m = 7.5 cm
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:
E = 0.01 J
Explanation:
Given that,
The mass of the cart, m = 0.15 kg
The force constant of the spring, k = 3.58 N/m
The amplitude of the oscillations, A = 7.5 cm = 0.075 m
We need to find the total mechanical energy of the system. It can be given by the formula as follows :
Put all the values,
So, the value of total mechanical energy is equal to 0.01 J.
Answer:
V(average)=6.37 V
Explanation:
Given Data
Peak Voltage=10V
Frequency=10 kHZ
To Find
Average Voltage
Solution
For this first we need to find Voltage peak to peak
So
Voltage (peak to peak)= 2× voltage peak
Voltage (peak to peak)= 2×10
Voltage (peak to peak)= 20 V
Now from Voltage (peak to peak) formula we can find the Average Voltage
So
Voltage (peak to peak)=π×V(average)
V(average)=Voltage (peak to peak)/π
V(average)=20/3.14
V(average)=6.37 V
Answer:
Answer is C. Both technicians A and B.
Refer below.
Explanation:
Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:
Both technicians A and B
