Internal energy is the sum of potential and kinetic energies in the particles of a substance.
The type of energy that refers to the sum of potential and kinetic energies in the particles of a substance is internal energy.
Internal energy is the total energy of all the particles in a substance, including the energy associated with their motion and stored energy due to their positions or arrangements. It consists of both potential energy (energy stored in the particles' positions) and kinetic energy (energy of the particles' motion).
For example, consider a gas in a container. The internal energy of the gas would be the sum of the potential energy of the gas particles due to their positions and the kinetic energy of the particles due to their motion.
Answer:
The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).
If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.
The correct option is option (1)
The faster movement of air on the upper surface of the paper creates lower pressure above the paper.
The movement of air is always from a region of higher pressure to a region of lower pressure.
As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.
So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.
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This is happened because "the air" above "moves faster" and "the pressure" is "lower".
Option: 1
Explanation:
Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.
Answer:
Explanation:
For refraction through a curved surface , the formula is as follows
μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m
u , object distance = - .465 m
1 / v + 1.333 / .465 = (1 -1.333 )/1.95
1 / v + 2.8667 = - .171
1 / v = - 2.8667 - .171 = - 3.0377
v = - .3292 m
= - 32.92 cm
image will be formed in water.
c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .
= (1.33 x 32.92) / (1 x 46.5)
= .94
size of image of teeth = .94 x 5
= 4.7 cm .
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm
The wooden block must be attached at the 57.6 cm mark on the meter stick.
We can solve this equation for the distance of the wooden block from the pivot point, which is the mark on the meter stick where it must be attached so that the meter stick balances horizontally when lowered into fresh water.
The density of aluminum is 2700 kg/m³ The volume of the meter stick is 0.01 m³ (length times width times thickness). The acceleration due to gravity is 9.81 m/s²
Substituting these values into the equation above, we get:
distance of wooden block from pivot point = -44.89 Nm / [(2700 kg/m³)(0.01 m³)(9.81 m/s²)]
= 0.576 m
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#SPJ3
Answer:
Stop cheating in exam
Explanation:
Shame!!!!
I am sorry but I will have to refer you to the student conduct at UTA.
Answer:
a. 3.73 m/s b. 27.8 m/s²
Explanation:
(a) Calculate his velocity (in m/s) when he leaves the floor.
Using the conservation of energy principles,
Potential energy gained by basketball player = kinetic energy loss of basket ball player
So, ΔU + ΔK = 0
ΔU = -ΔK
mg(h' - h) = -1/2m(v'² - v²)
g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor
Substituting the values of the variables into the equation, we have
9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)
9.8 m/s²(0.710 m) = -1/2(-v²)
6.958 m²/s² = v²/2
v² = 2 × 6.958 m²/s²
v² = 13.916 m²/s²
v = √(13.916 m²/s²)
v = 3.73 m/s
(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.
Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m
So, making, a subject of the formula, we have
a = (v² - u²)/2s
Substituting the values of the variables into the equation, we have
a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)
a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)
a = 13.913 m²/s²/(0.50 m)
a = 27.83 m/s²
a ≅ 27.8 m/s²
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.
Answer:
A) A₁ V₁ = A₂V₂
B) V₂ = 19 m /s
C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg
D) P₂ = 1.88 atm
Explanation:
A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;
A₁ V₁ = A₂V₂
Where;
A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.
B) Using the formula given in A above, we obtain;
π x 1.2² x 4.75 = π x 0.6² x V₂
V₂ x 0.36 = 6.84
V₂ = 6.84/0.36
V₂ = 19 m /s
c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;
P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg
Where;
P₁ and P₂ are pressure at ground and second floor respectively
v₁ and v₂ are velocity at ground and second floor respectively
h₁ and h₂ are height at ground and second floor respectively ρ is density of water.
Thus, plugging in the relevant values to obtain;
4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)
(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10
P₂ = 1.9 X 10⁵ N/m² = 1.88 atm