Answer:
Reverse Osmosis
Explanation:
Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.
Answer:
filtration is used to separate
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N
Answer:
7.46×10⁻⁶ m
Explanation:
Applying,
F = kqq'/r²............ Equation 1
make r the subject of the equation
r = √(F/kqq').......... Equation 2
From the question,
Given: F = 8 N, q' = q= 4 C
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 2
r = √[8/(4×4×8.98×10⁹)]
r = √(55.7×10⁻¹²)
r = 7.46×10⁻⁶ m
A block of mass 10kg is suspendet at a diameter of 20cm from the centre of a uniform bar im long, what force is required to balance it at its centre of gravity by applying the fore at the other end of the bar?
Answer:
4 kg of force
Explanation:
Force = (mass x distance to fulcrum) / length of fulcrum to end
Subsitute values
F = (10 x 20)/50
F =4
A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and 1510 N, respectively. The acceleration of the cable is 0.620 m s 2 , upward. What is the tension in the cable (a) below the worker and ( b) above the worker
Answer:
Explanation:
a)
Below the worker , the tension in cable is pulling the crate . Let the tension be T₁ .
weight of crate is acting downwards .
Total weight 1510 N.
Net force acting on both = T₁ - 1510
Applying second law of Newton ,
T₁ - 1510 = 1510 / 9.8 x 0.62 [ 1510 / 9.8 = mass of crate ]
T₁ - 1510 = 95.5
T₁ = 1605.5 N.
b )
Above the worker , the tension in cable is pulling both the worker and the crate . Let the tension be T₂ .
weight of both worker and crate is acting downwards .
Total weight = 965 + 1510 = 2475 N.
Net force acting on both = T₂ - 2475
Applying second law of Newton ,
T₂ - 2475 = 2475 / 9.8 x 0.62 [ 2475 / 9.8 = mass of both worker and crate ]
T₂ - 2475 = 156.6
T₂ = 2631.6 N.
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb
Answer: [tex]29.85\ W/m^2[/tex]
Explanation:
Given
Power [tex]P=60\ W[/tex]
Distance from the light source [tex]r=0.4\ m[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Inserting values
[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]
Answer:
29.85 W/ m^2
Explanation:
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
convert 2.4 milimetre into metre
Answer is 0.0024
Explanation
divide the length value by 1000.
what is the force of a body which have mass of 7 kg
Answer:
Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N
Answer:
10 m/s2 or 70 newtons.
Explanation:
............................
............
An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.
Answer:
Option (d).
Explanation:
According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.
So,
Let the pressure is P, volume is V and T is the absolute temperature of the gas.
Pressure proportional to the reciprocal of the volume.
[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]
The correct option is (d).
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
Here,
v = 45m/s
t = 5s
d = v × t
Therefore,
d = 45 × 5
= 225m
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 54000 m/s. The experiment is repeated with a He+ ion (charge e, mass 4 u).What is the ion's speed at the negative plate?
An elevator with its occupants weighs 2400 N and is supported by a vertical cable. What is the tension in the cable if the elevator is moving up with its speed decreasing at a rate of 1.7
Answer:
Hope you find it useful. please correct me if I am wrong
The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.
What is tension?Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.
Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.
Given the force due to the weight of the elevator = mg = 2400N
m = 2400/9.8 Kg
The elevator deaccelerating while moving upward, a = -1.7 m/s²
According to Newton's 3rd law: T - mg = ma
T - 2400 = (2400/9.8) × (-1.7)
T = 2400 - 416.32
T = 1983.67 N
Learn more about tension, here:
brainly.com/question/28965515
#SPJ5
Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.
b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]
Answer:
wareffctgggyyggghhhh
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
What are the differences among elements, compounds, and mixtures?
Answer:
Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.
••••••••••••••••
Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.
•••••••••••••••
A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements.
------------------------------
Hope it helps...
Have a great day!!!
Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.
A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?
Answer:
The exit velocity of water is B. 15 m/s.
Explanation:
According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.
If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity is v₂, then,
Rewrite the expression for v₂.
Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.
The exit speed of water from the hose is 15 m/s.
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]
The second equation of work done is set up as follows;
[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]
solve equation (1) and equation (2) together;
[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]
[tex]l = 9 \ cm[/tex]
Therefore, the natural length of the spring is 9 cm
When you stand on tiptoes on a bathroom scale, there is an increase in
A) weight reading.
B) pressure on the scale, not registered as weight.
C) both weight and pressure on the scale.
D) none of the above
Answer:
B) Pressure on the scale, not registered as weight.
Explanation:
This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point
Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength.
Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10−4 m2, a typical value.)
Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area?
Answer:
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
Explanation:
Given:
cross sectional area of the bone, [tex]a=4.8 \times 10^{-4} ~m^2[/tex]
factor of up-scaling the dimensions, [tex]s=10[/tex]
Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.
The scaled up area is:
[tex]a_s=a\times s^2[/tex]
[tex]a_s=[4.8 \times 10^{-4}]\times 10^2[/tex]
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. The new cross-section area will be 4.8×10⁻² m².
What is the area?The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. Its unit is m².
Given data in the problem
a is the crossectional area of conical bone = 4.8×10⁻⁴m².
s is the factor of up-scaling the dimensions =10
For two degrees of dimension, the upscaled area will be square of the given area.
The scaled-up area will be
[tex]\rm a_s=a\times s^2\\\\ a_s= 4.8\times10^{-4}\times {10}^2\\\\\ \rm a_s=4.8\times10^{-2}\;m^2[/tex]
Hence the new cross-section area will be 4.8×10⁻² m².
To learn more about the area refer to the link;
https://brainly.com/question/1631786
A small plane tows a glider at constant speed and altitude. If the plane does 2.00 * 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal
Answer:
θ = 57.4°
Explanation:
The complete formula to find out the work done by the plane is as follows:
[tex]W = FdCos\theta[/tex]
where,
W = Work = 200000 J
F = Force = Tension = 2560 N
d = distance = 145 m
θ = angle between rope and horizontal = ?
Therefore,
[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]
θ = 57.4°
in what part of the plant is glucose suger made?
[tex]\large \mid \underline {\bf {{{\color{navy}{Leaf \: \: \: Chloroplast \: ...}}}}} \mid[/tex]
☛ More Information :Green plants manufacture glucose through a process that requires light, known as photosynthesis. Glucose is stored in the form of starch in plants.
If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?
Answer:
the current that flows through the component is 2.42 A
Explanation:
Given;
resistance of the electrical component, r = 53 Ω
the voltage of the source, V = 128 V
The current that flows through the component is calculated using Ohm's Law as demonstrated below;
[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]
Therefore, the current that flows through the component is 2.42 A
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
Answer:
Heat is flowing into the metal.
Explanation:
From the question given above, the following data were obtained:
Mass (M) of iron = 150 g
Initial temperature (T₁) = 25.0°C
Final temperature (T₂) = 73.3°C
Direction of heat flow =?
Next, we shall determine the change in the temperature of iron. This can be obtained as follow:
Initial temperature (T₁) = 25.0 °C
Final temperature (T₂) = 73.3 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 73.3 – 25
ΔT = 48.3 °C
Next, we shall determine the heat transfered. This can be obtained as follow:
Mass (M) of iron = 150 g
Change in temperature (ΔT) = 48.3 °C
Specific heat capacity (C) of iron = 0.450 J/gºC
Heat (Q) transfered =?
Q = MCΔT
Q = 150 × 0.450 × 48.3
Q = 3260.25 J
Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.
The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Answer:
The potential at a distance of 0.9 m is 266.67 V.
Explanation:
Charge = Q
Potential is 400 V at a distance 0.6 m .
Let the potential is V at a distance 0.9 m.
Use the formula of potential.
[tex]V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V[/tex]
Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
