Answer:
C
Explanation:
covalent. Hope that helps!
an electric iron draws a current of 5a when connected to a 250v supply. calculate the energy used buy the iron if it's used for 10 minutes
Answer:
The energy used by the iron is 0.2083 KWH.
Explanation:
Electric Energy
The energy consumed by an electrical appliance is given by:
E = P.t
Where:
P = Power consumed
t = Time the appliance is connected to the power supply
The power can be calculated as a function of the voltage V and the current I:
P = V.I
Therefore, the energy is:
E = V.I.t
The electric iron draws a current of I=5 A when connected to a V=250 V supply for t=10 minutes. The time must be expressed in hours, thus t=10/60=0.167 hours.
Thus, the energy is:
E = 250 V * 5 A * 0.167 H
E = 208.33 WH
Converting to KWH:
E = 208.33/1000 = 0.2083 KWH
The energy used by the iron is 0.2083 KWH
What are the factors that affect the size of an impact crater?
Answer:
The following factors could be responsible for size of an Impact Crater:
Geography of the region where it impacted.Size of the Impactor Velocity of the ImpactorHope it helps!<3
A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
Answer:
Pressure = Force/Area
347 = 1300/Area
Area = 1300/347
area= 3.47m2
Answer:
I hope the above pic will help you
I will always help you understanding your assingments have a great day
#Captainpower
a 9.0-kg dog runs at 4.0 m/s and jumps onto a stationary skateboard the mas of the skateboard is 1.0 m/s what speed is the speed of the skating dog?
Answer:
1.3
Explanation:
sbsbsbsnsnssnssssssiejejeneene
Calculate the change in entropy of 0.020 kg of ice when it melts at 0.0°C. The heat of fusion of ice is 3.36 x 10'J/kg.
Answer:
S = 2461.53 [kJ]
Explanation:
The change in entropy in a process such as melting can be calculated by means of the following expression.
[tex]S=\frac{H*m}{T}[/tex]
where:
S = entropy [kJ]
H = fusion heat = 3.36*10¹ [J/kg]
m = mass = 0.02 [kg]
T = temperature in kelvin = 273 [K]
[tex]S = \frac{0.02*3.36*10^{1} }{(273+0)}\\S = 2461.53 [kJ][/tex]
A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of the circle so that the cord remains taut and the ball’s path remains circular.
Answer:
The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.
Explanation:
By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:
[tex]K =U_{g}[/tex] (1)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]U_{g}[/tex] - Gravitational potential energy, measured in joules.
By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h[/tex]
[tex]v = \sqrt{2\cdot g\cdot h}[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]h[/tex] - Maximum height of the ball, measured in meters.
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]h = 5\,m[/tex], then the initial speed of the ball is:
[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]
[tex]v\approx 9.903\,\frac{m}{s}[/tex]
The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.
When a ball is thrown upward, there is a loss of ____________ energy, but a gain in ___________ energy.
Answer:
Explanation:
Loss of potential energy but a gain in kinetic (becasue its falling)
Answer:
There is a loss of potential energy and a gain of kinectic
Explanation:
What is the mechanical advantage of the inclined plane below?
a) 2.6
b) 0.8
c) 2.4
d) 0.6
Answer;
2.6
Explanation;
Formular for mechanical advantage is length of the slope divided by the height of the inclined plane.[tex]1.85 \div 0.72 = 2.56[/tex]Approximately 2.6A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?
Answer:
0.0158m/s
Explanation:
Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;
m1u1 + m2u2 = m1v1 + m2v1
m1 and m2 are the masses of the object
u1 and u2 are the initial velocities
v1 and v2 are the final velocities
Given
m1 = 300g = 0.3kg
u1 = 1.10m/s
m2 = 4.00kg
u2 = 0m/s (at rest)
v1 = 0.890
v2 = ?
Substitute the given values into the formula;
0.3(1.10) + 0 = 0.3(0.89) + 4v2
0.33 = 0.267 + 4v2
0.33-0.267 = 4v2
0.063 = 4v2
v2 = 0.063/4
v2 = 0.0158m/s
Hence the speed of the large cart after the collision is 0.0158m/s
According to the question:
Mass,
[tex]m_1 = 300 \ g = 0.3 \ kg[/tex][tex]m_2 = 4.00 \ kg[/tex]Final velocity,
[tex]u_1 = 1.10 \ m/s[/tex][tex]u_2 = 0 \ m/s[/tex]Initial velocity,
[tex]v_1 = 0.890 \ m/s[/tex][tex]v_2 = \ ?[/tex]By using the law of conservation, we get
→ [tex]m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1[/tex]
By substituting the values, we get
→ [tex]0.3(1.10)+0 = 0.3(0.89)+4v_2[/tex]
→ [tex]0.33=0.267+4 v_2[/tex]
[tex]0.33-0.267 = 4 v_2[/tex]
[tex]0.063 =4v_2[/tex]
[tex]v_2 = \frac{0.063}{4}[/tex]
[tex]= 0.0158 \ m/s[/tex]
Thus the response above is appropriate.
Learn more about speed here:
https://brainly.com/question/14478313
Explain how polarization of a cell increases the cell's internal resistance.
(2
2.
Answer:
Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.
A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.
One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.
A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.
LOCAL ACTION
When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.
Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.
The velocity of a wave is 500 m/s . The wavelength of the wave is 3.2 m . What is the frequency of the wave ?
Answer:
f=156.25Hz
Explanation:
v=fT
500=fx3.2
500=3.2f
divide both sides by 3.2
f=156.25Hz
10) The coil of transformer is coated by enamel why
Answer:
Enamel is used to coat the wires, it is the thinnest possible insulator. The coils are made up of large number of turns and enamel makes it possible to cram a lot of wires (coils) in much smaller space.
In a step-down transformer the secondary coil is made of thick copper coil.
Explain
(2 marks)
Answer:
In a step-up transformer, the secondary coil is made of same wire as the primary coil but it is rolled with the presence of the insulated which never conduct the opposite current to the coil.
Hence it function property without meeting any fire and other problem.
Even if you want to get clear ideas, then you have to go with official website.
A car starting at rest accelerates to a speed of 6miles/hr in 5 sec what is the average acceleration
Answer:
1.2 mi/hr/sec
Explanation:
a = Vf - Vi / t (formula)
-----------------------------------
6mi/hr - 0 mi/hr 6 mi/hr
a = ____________ = _______ = 1.2 mi/hr/sec
5 sec 5 sec
What makes metal a good conductor?
Answer:
It is one of the strongest materials in the world and can hold anything
What is the volume of a brick that is 30 cm long, 8 cm wide, and 10 cm tall?
help please Derive an equation
Ta=1.44T1/2
Answer:
To derive an equation you must indicate the variable you want to solve for.Here we have tension of an object A and Tension 1.Two variables or unknown are given hence we cannot derive any other equations.I dont understand physics.
Answer:
same but from what i know is the newtons thingy
Explanation:
hope that helps!
A cross-country skier travels 40 meters north in 2 minutes. What is the skiers velocity?
Answer:
v = 0.33 [m/s]
Explanation:
We must remember that speed is defined as the relationship between the displacement in a given time. In this way, we can propose the following equation.
[tex]v=d/t[/tex]
where:
v = velocity [m/s]
d = displacement = 40 [m]
t = 2 [min] = 120 [s]
Now replacing we have:
[tex]v=40/120\\v=0.33[m/s][/tex]
The direction of electrical force is based on the fact that like charges will
yes
Explanation:
because hindi ko tlaga ala
Answer:
the answer is repel
Explanation:
Car 2 and Car 3 are driving down the road at the same speed. Car 1, going 15 m/s, has just passed Car 2 and is approaching Car 3 when he sees Person A standing in front of him. The driver of Car 1 quickly honks his horn as a warning, but does not brake. At the moment that Car 1 honks his horn, Person B is riding his bicycle, behind, but in the same direction as Car 1 at 7 m/s. Which person hears the highest frequency from the horn?
Answer:
Person A
Explanation:
Person A hears the highest frequency from the horn than Person B because Person A is present in front of a Car 1 while the Person B present behind the Car 1. The car horn is present at the front so the Person A will hear the highest frequency from the horn as compared to Person B which hears lower frequency from the horn. If the horn is present at the back side so Person B will hear highest frequency from the horn.
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
Answer:
c. because A will land first becuase its heavier :)
Explanation:
list at least three examples of circular motion
Answer:
When an object moves with a constant speed in a circular path, then its motion is called as Circular motion.
Examples of Uniform Circular Motion :--
The moon moves in uniform circular motion around the earth.
A stoned tied to a thread and rotated in circular motion.
An athlete running on a circular track.
A satellite revolving around Earth.
a 2kg block is attached to a horizontal ideal spring with a spring constant of 200 Newton per minute. when the spring has its equillibrium length of block is given a speed of 5mps. what is the maximum elongation of the spring
Answer:
The maximum elongation of the spring is 0.5 meters.
Explanation:
The statement is incorrect. The correct form is:
A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200 newton per meter. When the spring has its equillibrium length, the block has a speed of 5 meters per second. What is the maximum elongation of the spring?
The block experiments a simple harmonic motion, where there are no non-conservative forces and the total energy is the sum of translational kinetic energy of the mass and the elastic potential energy of the spring. The maximum elongation of the spring is done when elastic potential energy reach its maximum. By the Principle of Energy Conservation, the maximum elastic potential energy is equal to the maximum translational kinetic energy, which corresponds to the instant when the mass reaches the equilibrium position. Then, the equation modelling the system is:
[tex]U_{max} = K_{max}[/tex] (1)
Where:
[tex]U_{max}[/tex] - Maximum elastic potential energy of the spring, measured in joules.
[tex]K_{max}[/tex] - Maximum translational kinetic energy of the mass, measured in joules.
By definitions of the maximum elastic potential energy of the spring and the maximum translational kinetic energy of the mass, the expression above is expanded and simplified:
[tex]\frac{1}{2}\cdot k\cdot x_{max}^{2} = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]
[tex]x_{max} = \sqrt{\frac{m}{k} }\cdot v_{max}[/tex] (2)
Where:
[tex]x_{max}[/tex] - Maximum elongation of the spring, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{max}[/tex] - Maximum speed of the mass, measured in meters per second.
If we know that [tex]m = 2\,kg[/tex], [tex]k = 200\,\frac{N}{m}[/tex] and [tex]v_{max} = 5\,\frac{m}{s}[/tex], then the maximum elongation of the spring is:
[tex]x_{max} = \sqrt{\frac{2\,kg}{200\,\frac{N}{m} } }\cdot \left(5\,\frac{m}{s} \right)[/tex]
[tex]x_{max} = 0.5\,m[/tex]
The maximum elongation of the spring is 0.5 meters.
help please Derive an equation
Ta=1.44T1/2
Explanation:
To derive an equation you must indicate the variable you want to solve for.
Here we have tension of an object A and Tension 1.
Two variables or unknown are given hence we cannot derive any other equations.
3. A cart (m= 10.0 kg) is currently traveling with a velocity of 3.0 m/s. The cart then gains speed,
achieving a new velocity of 7.0 m/s after 8.0 seconds. Calculate the magnitude of the net force acting
on the cart.
A 0.0 N
B. 1.25 N
C. 5.0 N
D. 80N
First we calculate the acceleration of the cart:
a = (v₂ - v₁) / (t)
a = (7 m/s - 3 m/s) / (8 s)
a = 4 m/s / 8 s
a = 0.5 m/s²
Using the Newtons Second Law:
F = m×a
F = (10 kg)(0.5 m/s²)
F = 5 N
The correct option is CIf the body is at rest while I am pushing it with a force of 20 Newtons and the force of static friction is 20 Newtons, will it move or should I push with a force greater than 20
Answer:
It will not move, so you should push with a force greater than 20 N.
Explanation:
Static friction is a force that keeps a body at rest. This means static friction will be the force that keeps the body from moving when there is no push or pull hence you should overcome it with enough opposing force for the body to move.
In this case for the body to move , you should push with a force greater than 20 N for it to move.
A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 61 cm behind the slits is 6.2 mm What is the wavelength of the light used in this experiment?
Answer:
[tex]\lambda =533.6 nm [/tex]
Explanation:
the slits spacing, d = 0.21 mm
distance of screen, D = 61 cm
The condition for minima is given as
[tex]dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda[/tex]
So, first minima, n = 0
[tex]dsin(\theta_1) = \frac{1}{2}\lambda[/tex]
fifth minima, n = 4
[tex]dsin(\theta_5) = \frac{9}{2}\lambda[/tex]
[tex]d(sin(\theta_5) -sin(\theta_1))= 4\lambda[/tex]
For small angle
[tex]d(tan(\theta_5) -tan(\theta_1))= 4\lambda[/tex]
From the figure:
[tex]d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda[/tex]
[tex]\frac{d}{D} (y_5-y_1) = 4\lambda[/tex]
[tex]\lambda = \frac{d}{4D} (y_5-y_1)[/tex]
[tex]\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})[/tex]
[tex]\lambda =533.6 nm[/tex]
"533.5 nm" would be the wavelength of the light used in this experiment.
According to the question,
Slits spacing,
d = 0.21 mmDistance of screen,
D = 61 cmThe condition for minima is given as:
→ [tex]d (sin \Theta) = (n+\frac{1}{2} ) \lambda[/tex]
So,
1st minima, n = 0
→ [tex]d sin(\Theta_1)= \frac{1}{2} \lambda[/tex]
5th minima, n = 4
→ [tex]d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda[/tex]
For small angle,
→ [tex]d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda[/tex]
From the figure, we get
→ [tex]d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda[/tex]
[tex]\frac{d}{D} (y_5 -y_1) = 4 \lambda[/tex]
[tex]\lambda = \frac{d}{4D}(y_5-y_1)[/tex]
[tex]= \frac{0.021}{4\times 61} (6.2\times 10^{-3})[/tex]
[tex]= 533.6 \ nm[/tex]
Thus the above answer is correct.
Learn more:
https://brainly.com/question/14294313
B. How does potential energy change when
mass is increased?
Answer:
with the increase in mass of the object , the force of gravitation increases that eventually increases the potential energy if the object is still fixed in its relative position with surrounding objects.
Explanation:
hopes this will help you
. A pendulum of length l = 9.8 m hangs in equilibrium and is then given velocity
v
=0.2
m/s at its lowest point. What is the amplitude of the subsequent oscillation?
Answer:
the amplitude of the sequence oscillation=
ASO = length × velocity
= 9.8 × 0.2 = 19.6
ASO = 19.6