1. What happens to the current in a series circuit as it moves through each component? a. The current stays the same throughout the circuit.
b. The current will increase or decrease depending on the resistance.
c. The current decreases with each component it goes through.
d. The current increases with each component it goes through.
2. A 10-volt power supply is placed in series with two 5-ohm resistors. What is the current in the circuit after it passes through each of the two resistors?(1 point)
a. The current will stay the same at 1 amp after passing through both resistors.
b. The current will drop to 2 amps after the first resistor and then to 1 amp after the second resistor.
c. The current will stay the same at 2 amps after passing through both resistors.
d. The current will drop to 1 amp after the first resistor and then to 0 amps after the second resistor.
3. What is the voltage that passes through R1 and R2?
a. R1: 12 V, R2: 24 V
b. R1: 8 V, R2: 4 V
c. R1: 12 V, R2: 12 V
d. R1: 6 V, R2: 6 V
4. Which of the following correctly describes the magnitude of currents I1 and I2 ?
a. I1 is equal to I2
b. I1 and I2 approach zero
c. I1 is greater than I2
d. I1 is less than I2
5. If the energy of an electric charge flowing in a circuit is conserved, which of the following obeys the Kirchhoff junction rule?
a. The sum of the current flowing in is greater than the sum of current flowing out.
b. The sum of the current flowing in is less than the sum of the current flowing out.
c. The sum of the current flowing in is equal to the sum of current flowing out.
d. The sum of the current flowing in is zero and the sum of the current flowing out is greater than zero.
Answers
Answer: sorry here’s the answers, I didn’t feel like typing it all
Explanation:
The correct answer to the 5 questions are;
1) Option A; The current stays the same throughout the circuit.
2) Option A; The current will stay the same at 1 amp after passing through both resistors.
3) Option C; R1: 12 V, R2: 12 V
4) Option A; I1 is equal to I2
5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.
1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.
2) Formula for current is;
I = V/R
We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.
Thus; Current = 10/10 = 1 A.
The current will stay same at 1 A after passing through both resistors.
3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.
Option C is correct
4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.
5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.
Read more at; https://brainly.com/question/15394172
Related Questions
The human circulatory system is closed-that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart’s four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. If the aorta (diameter dad a) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? (a) da−−√ d a ; (b) da/2–√d a / 2 ; (c) 2da2d a ; (d) da/2d a /2.
Answers
The question is missing parts. The complete question is here.
The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?
(a) [tex]\sqrt{d_{a}}[/tex]
(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
(c) [tex]2d_{a}[/tex]
(d) [tex]\frac{d_{a}}{2}[/tex]
Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]
Explanation: The cross-sectional area of a vessel is a circle. So area is:
[tex]A=\pi.r^{2}[/tex]
Radius is half of a diameter, i.e.:
[tex]r=\frac{d}{2}[/tex]
Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:
[tex]r_{b}=\frac{d_{b}}{2}[/tex]
Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:
[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]
[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]
Area of aorta is
[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]
[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]
Area of aorta is equal the combined area of the branches, then:
[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]
Rearraging:
[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]
[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]
The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the
ground (or right before)?
Answers
Answer:
The final speed of the second package is twice as much as the final speed of the first package.
Explanation:
Free Fall Motion
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
[tex]v=gt[/tex]
And the distance traveled downwards is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]
Replacing into the first equation:
[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]
Rationalizing:
[tex]\displaystyle v=\sqrt{2gy}[/tex]
Let's call v1 the final speed of the package dropped from a height H. Thus:
[tex]\displaystyle v_1=\sqrt{2gH}[/tex]
Let v2 be the final speed of the package dropped from a height 4H. Thus:
[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]
Taking out the square root of 4:
[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]
Dividing v2/v1 we can compare the final speeds:
[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]
Simplifying:
[tex]\displaystyle v_2/v_1=2[/tex]
The final speed of the second package is twice as much as the final speed of the first package.
When you shock your self because of a door knob this is an example of
A.the law of conservation of charge
B. Force field
C.static discharge
D.torture lol
Answers
Answer:
C
Explanation:
Static electricity is what makes your hair stand up when you rub a balloon against it or gives you a shock from your doorknob. In static electricity, electrons are moved around mechanically (i.e. by someone rubbing two things together).
HOPE THIS HELPED AND LETTER D MADE ME LAUGH
How much distance did this object travel in meters between 0 and 10 seconds
Answers
Answer:
5m
Explanation:
Answer:
5
Explanation:
because that's the average of how far it when the most
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff 132 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) J (b) Suppose the projectile is traveling 89.3 m/s at its maximum height of y = 297 m. How much work has been done on the projectile by air friction? J (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
Answers
Answer:
a) The initial total mechanical energy of the projectile is 498556.296 joules.
b) The work done on the projectile by air friction is 125960.4 joules.
c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Explanation:
a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:
[tex]E = U_{g} + K[/tex] (Eq. 1)
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.
[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.
If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:
[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 498556.296\,J[/tex]
The initial total mechanical energy of the projectile is 498556.296 joules.
b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:
[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)
Where:
[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.
[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]
[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.
[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.
If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:
[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]
[tex]W_{loss} = 125960.4\,J[/tex]
The work done on the projectile by air friction is 125960.4 joules.
c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:
[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)
[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]
Where:
[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.
[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.
[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)
[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]
[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]
[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]
If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:
[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]
[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]
The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
A hawk is flying due west at 10.1 m/s carrying a field mouse in its talons. The mouse manages to break free at a height of 15 m. What is the magnitude of the mouse's velocity as it reaches the ground? Note: neglect air resistance
Answers
Answer:
When the mouse broke free from the hawk, its vertical velocity is zero since the mouse just fell from the grips of the hawk
We are given:
initial vertical velocity (v) = 0 m/s
initial horizontal velocity (u) = 10.1 m/s
height from the ground (h) = 15 m
acceleration due to gravity (a) = 10 m/s/s
final vertical velocity (vf) = v m/s
Solving for Final vertical velocity:
from the third equation of motion
(vf)² - (v)² = 2ah
replacing the variables
v² - 0² = 2(10)(15)
v² = 300
v = √300
v = 10√3 OR 17.3 m/s
43. The particle consists of fast moving electrons is.
Answers
Answer:
Explanation: someone help me please you want free points right
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.81 g, what is the tension in the web?
N=
Answers
Answer:
The tension in the web is 0.017738 N
Explanation:
Net Force
The net force exerted on an object is the sum of the vectors of each individual force applied to an object.
If the net force equals 0, then the object is at rest or moving at a constant speed.
The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.
A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:
T=W=m.g
The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:
[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]
[tex]T=0.017738\ N[/tex]
The tension in the web is 0.017738 N
Which statement describes the interaction between the north and south poles of two magents?
Answers
Answer:
A south pole attracts a north pole :)
Explanation:
What is the displacement from the forest to the doctor’s office?
Answers
Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.
procedure for determining the thermal conductivity of a solid involves embedding thermocouple in a thick slab of the material and measuring the response to a prescribed change in temperature at one surface. Consider an arrangement for which the thermocouple is embedded 10 mm from a surface that is suddenly brought to a temperature of 100degreeC by exposure to boiling water. If the initial temperature of the slab was 30degreeC and the thermocouple measures a temperature of 65degreeC, 2 minutes after the surface is brought to 100degreeC, what is the thermal conductivity. The density of the material is 2200 kg/m3 and the specific heat is 700 J/M- Find: What is the thermal conductivity of the material
Answers
Answer:
The thermal conductivity [tex]k = 1.4094 W/ m\cdot K[/tex]
Explanation:
From the question we are told that
The depth of the thermocouple from the surface is x = 10 mm = 0.01 m
The temperature is [tex]T_f = 100 ^o C[/tex]
The initial temperature is [tex]T_i = 30 ^o C[/tex]
The temperature of the thermocouple after t = 2 minutes( 2 * 60 = 120 \ seconds) is [tex]T_t = 65 ^o C[/tex]
The density of the material is [tex]\rho = 2200 kg/m^3[/tex]
The specific heat of the solid [tex]c_s = 700 J/kg \cdot K[/tex]
Generally the equation for semi -infinite medium is mathematically as
[tex]\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ][/tex]
Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]
[tex]\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}[/tex] this is from the Gaussian function table
[tex]0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )[/tex]
=> [tex]\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }[/tex]
=> [tex]\alpha = 9.1525 *10^{-7} \ m^2 /s[/tex]
Generally the thermal conductivity is mathematically represented as
[tex]k = \alpha * \rho * c_s[/tex]
[tex]k = 9.1525 *10^{-7} * 2200 * 700[/tex]
[tex]k = 1.4094 W/ m\cdot K[/tex]
11. What is the momentum of a truck with a mass of 3000 kg traveling north at a
speed of 25 m/s?
Answers
Answer:
We are given:
mass of the truck (m) = 3000 kg
velocity of the truck (v) = 25 m/s
Calculating the Momentum:
We know that the momentum of an object is equal to its mass times it's velocity
P = mv (where P is the momentum of the truck)
Momentum of the truck:
replacing with the values of the truck
P = mv
P = 3000 * 25
P = 75000 kg m/s
Therefore, the momentum of the truck is 75000 kg m/s
Which of the following elements are most likely to have similar chemical properties?
Answers
Answer:
transition metal, and inner transition metals groups are numbered 1-18 from left to right
Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what the magnitude of bigger vector?
Answers
Answer:
Vy = 80.5 [m/s]
Explanation:
In order to solve this problem we must use the Pythagorean theorem.
V = 90 [m/s]
The components are Vx and Vy:
Therefore:
[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]
where:
Vy = 2*Vx ; because one is twice of the other.
[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]
and the bigger vector is:
Vy = 40.25*2
Vy = 80.5 [m/s]
A mass of 15 kg is resting on a horizontal, frictionless surface. Force 1 of 206 N is applied to it at some angle above the horizontal, force 2 has a magnitude of 144 N and is applied vertically downward, force 3 has a magnitude of 5 N and is applied vertically upwards, and force 4 has a magnitude of 42 N and is applied in the -x direction to the object. When these forces are applied to the object, the object is moving at 20 m/s in the x direction in a time of 3 seconds. What is the normal force acting on the mass in Newtons
Answers
Answer:
N = 136.77 N
Explanation:
This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.
Let's use trigonometry to decompose the force F1
cos θ = F₁ₓ / F₁
sin θ = F_{1y} / F₁
F₁ₓ = F₁ cos θ
F_{1y} = F₁ sin θ
now let's apply Newton's second law to each axis
X axis
F₁ₓ - F4 = m a
Y axis
N + F3 + F_{1y} -F₂ -W = 0
the acceleration can be calculated with kinematics
v = v₀ + a t
since the object starts from rest, the initial velocity is zero v₀ = 0
a = v / t
a = 20/3
a = 6.667 m / s²
we substitute in the equation
F₁ₓ = F₄ + m a
F₁ₓ = 42 + 15 6,667
F₁ₓ = 142 N
F₁ cos θ = 142
cos θ = 142/206 = 0.6893
θ = cos⁻¹ 0.6893
θ = 46.42º
now let's work the y axis
N = W + F₂ - F₃ - F_{1y}
N = 15 9.8 + 144 -5 - 206 sin 46.42
N = 286 - 149.23
N = 136.77 N
PLEASEEEEE!!!!
A cube has a mass of 100 g and a volume of 50 mL. What is the density of the cube?
Answers
Answer:
The answer is 2.0 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\[/tex]
From the question
mass = 100 g
volume = 50 mL
We have
[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]
We have the final answer as
2.0 g/mLHope this helps you
distinguish between current and current density.
Answers
Answer:
1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor
Julie takes a cold glass of dear, doness houd out the register and pusi on the babe hathewwad warme up to room temperature, Julie sees buthles of gas form in the und and me to the face Which conclusion is best supported by Jules onervations? The liquid is a pure sutrance because it is closers O The liquid is a mixture because its temperature res The liquid is a pure substance because it remains lowd when The liquid is a mixture because it has a gas dissolved in a loud
Answers
Answer:
D. The liquid is a mixture because it has a gas dissolved in a liquid.
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside.
(a) Under this assumption, show that the cost of operating an air con- ditioner is proportional to the square of the temperature difference.
(b) Give a numerical example for a typical house and discuss implications for your electric bill.
(c) Suppose instead that heat enters the building at a rate proportional to the square-root of the temperature difference between inside and outside. How would the operating cost now depend on the temperature difference?
Answers
Answer:
Considering first question
Generally the coefficient of performance of the air condition is mathematically represented as
[tex]COP = \frac{T_i}{T_o - T_i}[/tex]
Here [tex]T_i[/tex] is the inside temperature
while [tex]T_o[/tex] is the outside temperature
What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity
So it implies that the air condition removes [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity
Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as
[tex]Q \ \alpha \ (T_o - T_i)[/tex]
=> [tex]Q= k (T_o - T_i)[/tex]
Here k is the constant of proportionality
So
since 1 unit of electricity removes [tex] \frac{T_i}{T_o - T_i}[/tex] amount of heat
E unit of electricity will remove [tex]Q= k (T_o - T_i)[/tex]
So
[tex]E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]
given that [tex]\frac{k}{T_i}[/tex] is constant
=> [tex]E \ \alpha \ (T_o - T_i)^2[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.
Considering the second question
Assuming that [tex]T_i = 30 ^oC[/tex]
and [tex]T_o = 40 ^oC[/tex]
Hence
[tex]E = K (T_o - T_i)^2[/tex]
Here K stand for a constant
So
[tex]E = K (40 - 30)^2[/tex]
=> [tex]E = 100K [/tex]
Now if the [tex]T_i = 20 ^oC[/tex]
Then
[tex]E = K (40 - 20)^2[/tex]
=> [tex]E = 400 \ K[/tex]
So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher
Considering the third question
Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside
We have that
[tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]
So
[tex]E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]
Assuming [tex]\frac{k}{T_i}[/tex] is a constant
Then
[tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.
5 13. If a train going 60 m/s hits the brakes, and it takes the train 2 minute 25 seconds to stop, what is the train's acceleration?
Answers
Answer:
Acceleration = -12/17 m/s ^2
Explanation:
Name the variables
V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)
A = acceleration( in this case, it is deceleration)
T = time ( time taken to fully stop)
Second Step: What equation should we use?
The 3 Big Motion in One Dimension Formulas are:
V = u+at
v^2 = u^2 + 2as
S = ut+ (1/2) at^2
For this question, we will be using the first one.
Third Step: Solve
V = 0 (the final velocity is 0, because the train stopped)
U = 60 (the train was going at 60 mps)
T = 85 seconds( 1 min and 25 sec)
0 = 60+ 85a
Use the first kinematic equation Vf = Vi + at. We are given everything here besides acceleration. We know Vf equals zero because it says the train comes to a stop.
0 = 60m/s + a*145s
-60m/s = 145a
a = -60/145 = -12/29 m/s^2
How long does it take a cheetah that runs with a velocity of 34m/s to run 750m?
Answers
Answer:
It would take 5 seconds
Explanation:
I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.
V = d / t
34 = 170 / t
34 x t = 170
34t = 170
t = 170 / 34
t = 5
What is the acceleration of a ball with a mass of 0.40 kg is hit with a force of 16N?
Answers
Answer:
40 m/s^2
Explanation:
Mass= 0.40 kg
Force= 16 N
Therefore the acceleration can be calculated as follows
F = ma
16= 0.40 × a
16= 0.40 a
a= 16/0.40
a= 40 m/s^2
Hence the acceleration is 40 m/s^2
Pest describes the act of using senses or tools to gather information?
ating a hypothesis
king an observation
mmarizing the results
ording the measurements
Answers
Answer:
Making an observation
Explanation:
The use of senses as a tool to gather information is described as making an observation.
While making an observation, the senses must be at alert.
Observation making is paramount to the scientific method. It is from observations that questions are asked and then hypothesis which can be tested are formulated. Observation can be carried out using the eyes, nose, feeling e.t.c. Nowadays, observation can also be carried out using some sophisticated equipment in the laboratory. This necessary for phenomenon the eludes our natural senses. A scientist must be a keen observer and their senses must be sharp.Therefore, the act of using senses or tools to gather information is called making an observation.
Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of this circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.”

(A) What was the initial position of the object?
(B) What is the final position of the object?
(C) What is the distance traveled by the object?
(D) What is the displacement of the object?
(E) Of the three underlined quantities, which are numerically equal?
Answers
A car rounds a banked curve as we will discuss in class on Tuesday. The radius of curvature of the road is R and the banking angle is θ. (a) In the absence of friction, what is the safe speed for the car to take this curve? (b) Now assume the coefficient of friction between the car’s tires and the road is µs. Determine the range of speeds the car can have without slipping up or down the road. (c) What is the minimum value of µs that makes the minimum speed zero? (d) If θ = 25.0 ◦ , for what values of µs can the curve be taken at any speed? Note: The upper limit of µs you will find is practically impossible to achieve for the car’s tires and the road.
Answers
Answer:
A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) µ_s = tan θ
D) µ_s = 0.4663
Explanation:
A) The forces acting on the car will be;
Force due to friction; F_f
Force due to Gravity; F_g
Normal Force; F_n
Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.
Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n
Thus, sum of forces about the vertical j^ direction gives;
ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0
Since F_f = µ_s × F_n ;
F_n•cos θ − mg + (µ_s × F_n × sin θ) =0
F_n = mg/[cos θ + (µ_s•sin θ)]
Also, sum of forces about the centre i^ direction gives;
ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r
Plugging in formula for F_n gives;
ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r
Making v the subject gives;
v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B) What we got in a above is the minimum speed the car can have while going round the turn.
The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.
Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;
v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
Thus the range is;
√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;
ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0
Thus;
mg(sin θ - µ_s•cos θ) = 0
Making µ_s the subject gives;
µ_s = sin θ/cos θ
µ_s = tan θ
D) If θ = 25.0°;
Thus;
µ_s = tan 25
µ_s = 0.4663
A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.
Answers
Answer:
The time for the change in the angular velocity to occur is 14.08 secs
Explanation:
From the question,
the angular acceleration is - 4.46 rad/s²
Angular acceleration is given by the formula below
[tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex]
Where [tex]\alpha[/tex] is the angular acceleration
[tex]\omega[/tex] is the final angular velocity
[tex]\omega _{o}[/tex] is the initial angular velocity
[tex]t[/tex] is the final time
[tex]t_{o}[/tex] is the initial time
From the question
[tex]\alpha[/tex] = - 4.46 rad/s²
[tex]\omega _{o}[/tex] = 0 rad/s (starting from rest)
[tex]\omega[/tex] = -31.4 rad/s
[tex]t_{o}[/tex] = 0 s
Now, we will determine t
From [tex]\alpha =\frac{\omega -\omega _{o} }{t - t_{o} }[/tex], then
[tex]-4.46 = \frac{-31.4 - 0}{t - 0}[/tex]
[tex]-4.46 = \frac{-31.4}{t}[/tex]
[tex]t = \frac{-31.4}{-4.46}[/tex]
t = 7.04 secs
This is the time spent in one direction,
Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t
Hence,
The time is 2×7.04 secs = 14.08 secs
This is the time for the change in the angular velocity to occur.
Chen is testing the friction of three surfaces. He pushes the same ball across three different surfaces with the same force and measures the distance the ball rolls over each surface. The ball moved ten inches across Surface 1, six inches across Surface 2, and fifteen inches across Surface 3. Which could most likely describe the three surfaces?
Answers
Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
Answer: its C
Explanation:
A 55-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 40.0o above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.0 m?
Answers
Answer:
478.75 J
Explanation:
W=force* displacement
constant speed= (a=0) net F=0
Horizontal component of tension
Tcosx
125Ncos40= 95.76 N
W= (95.76 N)(5 m)
=478.75 J
The work done in moving the crate across the given distance is 478.75 J.
The given parameters;
Mass of the packing create, m = 55 kgAngle of inclination of the rope, Ф = 40°Tension on the rope, T = 125 NDistance through which the crate is the moved, d = 5 mThe work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.
The work-done in moving the crate is calculated as;
W = Tcos(Ф) x d
W = 125cos(40) x 5
W = 478.75 J.
Thus, the work done in moving the crate across the given distance is 478.75 J.
Learn more here: https://brainly.com/question/19498865
What is the momentum of a 20.0 kg scooter traveling at 5.00 m/s?
Answers
Answer:
The answer is 100 kgm/sExplanation:
To find the momentum of an object given it's mass and velocity we use the formula
momentum = mass × velocityFrom the question
mass = 20 kg
velocity = 5 m/s
We have
momentum = 20 × 5
We have the final answer as
100 kgm/sHope this helps you
Sand is made of different types of rocks. The shapes and sizes are not all the same. Sand is a ----?
Answers
Answer:
A soil
Explanation:
Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.
Answer:
mixture
sand is a mixture