Answer:
weight is 3.50 x 10^5 N
force is 1.52 * 10^6 N
pressure is 1.25 * 10^5 Pa
Explanation:
given data
Given data
depth = 2.60 m
density = 915 kg/m3
length = 5.00 m
width = 3.00 m
to find out
weight of the olive oil, force of air pressure and the pressure exerted upward
solution
we know density = mass / volume
mass = density* width *length *depth
mass = (915)(3)(5)(2.60)
mass = 3.57 x 10^4 Kg
so weight = mg = 3.57 x 10^4 (9.81) = 3.50 x 10^5 N
weight is 3.50 x 10^5 N
and
we know force = pressure * area
area = 3 *5 = 15 m²
and we know atmospheric Pressure is about 1.01 * 10^5 Pa
so force = 1.01 * 10^5 (15) = 1.52 x 10^6 N
force is 1.52 * 10^6 N
and
we know Fup - Fdown = Weight
so
Fup = 1.52 * 10^6 + 3.50 * 10^5
Fup = 1.87 * 10^6 N
so pressure = Fup / area
pressure = 1.87 * 10^6 / 15
pressure is 1.25 * 10^5 Pa
Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.
Answer:
B. Energy increases with decreasing wavelength and increasing frequency.
Explanation:
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.
Answer:
option C
Explanation:
the correct answer is option C
When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.
According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.
This law's application is used in machines like hydraulic jacks.
Answer:
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
Explanation:
Given that
g₁ = 9.96 m/s²
g₂ = 9.72 m/s²
The actual value of g = 9.8 m/s²
a)
The difference Δ g = 9.96 -9.72 =0.24 m/s²
b)
For first one :
For second :
c)
The mean g(mean )
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.
In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:
|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%
The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:
For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%
For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%
The percent error of the mean involves doing the above but using the mean of the experimental measurements:
|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%
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Answer:
The initial momentum of the ball is 8 kg-m/s.
Explanation:
Given that,
Mass of the ball is 4 kg
Initial speed of the ball is 2 m/s
Force applied to the ball is 5 N for 3 seconds
It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :
So, the initial momentum of the ball is 8 kg-m/s.
Answer:
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Explanation:
Whilst first impressions can be important, your body language during the interview can make or break your overall performance. With experts saying that between 75-90% of communication is non-verbal, it is important to think about what your body is saying about you during an interview.
Answer:
Image is virtual and formed on the same side as the object, 19.29 cm from the lens.
The height of the image is 0.40509 cm
Image is upright as the magnification is positive and smaller than the object.
Explanation:
u = Object distance = 100 cm
v = Image distance
f = Focal length = -23.9 cm (concave lens)
= Object height = 2.1 cm
Lens Equation
Image is virtual and formed on the same side as the object, 19.29 cm from the lens.
Magnification
The height of the image is 0.40509 cm
Image is upright as the magnification is positive and smaller than the object.