Answer:
the selected configuration is PARALLEL
Explanation:
There are two ways to connect electrical circuits.
In series in this type of circuit the equipment is connected to the same line, in this case the current of the circuit circulates through all the inputs is the same, but the voltage in each one is a part of the total voltage, this circuit has the problem that if an appliance is disconnected or damages the power and the entire circuit is interrupted
V_total = ∑ V_i
In a parallel circuit the equipment is connected in two two cables, for which the current is divided in each branch, the voltage in all the branches is the same, this type of circuit has the advantage that if one equipment is damaged or disconnects the others they can continue working
V_total = V
Therefore, in consequence of the above, the selected configuration is PARALLEL
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
A person weighs 60kg. Calculate the amount of work done if he is raised 12m into the air.
Answer: I belive the answer would be 720 Joules
Explanation:
Work can be calculated by multiplying an objects weight times the distance traveled. Work is measured in joules (J).
Formula: W x D = J
W= weight
D= Distance
J= joules of energy
Hope it helps
an airplane normally flying at 80 km/h north encounters a wind from the west of 10km/h at right angle to its foward motion a crosswind what will its resultant velocity be ?
Answer:
v = 80.62 km/h
Explanation:
Given that,
An airplane normally flying at 80 km/h north encounters a wind from the west of 10km/h at right angle to its forward motion a crosswind.
We need to find the resultant velocity of the airplane. It can be calculated as follows :
[tex]v=\sqrt{80^2+10^2} \\\\v=80.62\ km/h[/tex]
So, the required velocity is 80.62 km/h.
Assuming no friction, how does the initial gravitational potential energy of
the marble on a downward slope compare to the final kinetic energy?
a) they are the same
b) the initial gravitational potential energy is greater than the final kinetic energy
c) the initial gravitational potential energy is less then the final kinetic energy
Answer:
a) They are the same.
Explanation:
Assuming no friction, there should be no energy transfer and thus the Law of Conservation of Energy says:
[tex]PE=KE,\\mgh=\frac{1}{2}mv^2[/tex]
These types of problems also disregard any air resistance the surface of the object may cause. Therefore, no energy is transferred and from the Law of Conservation of Energy, [tex]100\%[/tex] of energy is preserved.
explain resolution of Force
Answer:
it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.
Explanation:
2) Given R = 3 ohms and R, = 1 ohm and V = 12 volts
I
a) Find the total resistance.
b) Find the current in the circuit:
c) Find the voltage drop in each resistor:
Answer:
a) because this is in series, we have:
the total resistance is 3 + 1 = 4 (ohm)
b) the curren in the circuit is 12/4 = 3 (A)
c) the voltage in R = 3 ohm is 3.3 = 9 (V)
the voltage in R = 1 ohm is 12 - 9 = 3 (V)
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 sWhich action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
friction?
a. Static, Rolling, Sliding
b. Sliding, Rolling, Static
c. Rolling, Static, Sliding
d. Static, Sliding, Rolling
Answer:
[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]
Explanation:
Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.
During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.
Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
find the volume of an object with a density of 3.2 g/mL and a mass of 12 g.
Q1. A man wants to install a surveillance mirror in his shop, which mirror should he use?(1)
a) Convex mirror
b) Concave mirror
c) Plane mirror
d) Both (a) and (b)
answer is convex mirror
Explanation:
A
Because convex mirror will provide maximum view
The force of gravity acting on an object is directed through this
center of gravity and toward the center of the
Answer:
Earth.
Explanation:
Center of gravity can be defined as the specific point where all of the weight of an object is concentrated.
Generally, all the objects found around the world all have a center of gravity.
When an object is balanced so that a displacement lowers its center of gravity, the object is said to be in stable equilibrium.
Hence, the force of gravity acting on an object is directed through this center of gravity and toward the center of the earth.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, weight is given by the formula;
[tex] Weight, W = mg [/tex]
Where;
m is the mass of an object.
g is acceleration due to gravity.
Kiara starts at 4, walks 6 blocks left and 2 blocks right. What is her displacement?
describe measurement in our daily life
. Dan drags a box across the floor. He uses 95 N of force and moves the box 12 m. How much work does he do? 0.13 J 1,140 J 7.9 J 107 J
Answer:
[tex]1140\:\mathrm{J}[/tex]
Explanation:
Work is given by [tex]W=F\Delta x[/tex], where [tex]F[/tex] is force and [tex]\Delta x[/tex] is displacement.
Plugging in given values, we get:
[tex]W=95\:\mathrm{N}\cdot12\:\mathrm{m}=\fbox{$1140\:\mathrm{J}$}[/tex].
Explain how momentum is determined and conserved.
ASAP!!
Explanation:
Momentum is conserved in the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system.
Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
help me help me help me
In the equation for the gravitational force between two objects, which quantity must be squared?
•mi
•m2
•G
•d
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
Therefore, the quantity to be squared is dThe formula is given as:
Fg = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]
So d is the quantity that must be squared
3. As the mass of an object increases, the force of gravity
Answer:
As the mass of an object increases, the force of gravity increases as well.
Explanation:
Objects with more mass have more gravity. They work together.
On a slope where does a marble have to most kinetic energy?
a) it is always the same
b) at the initial position
c) at the final position
d) somewhere between the initial and the final position
Answer:
C
Explanation:
Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.
What relationship must exist between an applied force and the velocity of a moving object if uniform circular motion is to result?
Answer:
See explanation
Explanation:
Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.
The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.
So we can write;
F = mv^2/r
calculate the average speed of talias car during the trip
Answer:
We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any other guy
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching but you're too shy to say it
Inside we both know what's been going on
We know the game and we're gonna play it
And if you ask me how I'm feeling
Don't tell me you're too blind to see
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
Never, ever desert you
We've known each other for so long
Your heart's been aching but
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
I'll never, ever desert you
Explanation:
RICK ROLLED
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s. Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 23.1 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.570 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.0 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 21.3 m.
1. How much time is required for the car to pass the truck?
2. What distance does the car travel during this time?
3. What is the final speed of the car?
Answer:
16.16 s
381.5 m
28.21 m/s
Explanation:
Acceleration of the car = [tex]0.57\ \text{m/s}^2[/tex]
The distance the car moves is [tex]23.1+25+21.3+5=74.4\ \text{m}[/tex]
Initial position of the car is [tex]23.1+21.3+5=49.4\ \text{m}[/tex]
Position of the truck is given by the equation
[tex]x_r(t)=49.4+19t[/tex]
Position of the car is given by the equation
[tex]x_c(t)=19t+\dfrac{1}{2}\times0.57t^2[/tex]
Difference in their positions is 25 m
[tex]x_c(t)-x_r(t)=25\\\Rightarrow 19t+\dfrac{1}{2}\times0.57t^2-(49.4+19t)=25\\\Rightarrow 0.285t^2-74.4=0\\\Rightarrow t=\sqrt{\dfrac{74.4}{0.285}}\\\Rightarrow t=16.16\ \text{s}[/tex]
Time required for the car to pass the truck is 16.16 s
[tex]x_c(16.16)=19\times 16.16+\dfrac{1}{2}\times0.57\times 16.16^2=381.5\ \text{m}[/tex]
The distance the car traveled during this time is 381.5 m
[tex]v_c(16.16)=19+0.57\times 16.16=28.21\ \text{m}[/tex]
The final speed of the car is 28.21 m/s