Answer:
ejkshdkashalsflasfaksg
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular acceleration(assuming it was constant)
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
Consider the video tutorial you just watched. Suppose we repeat the experiment, but this time place the divider closer to one side of the tube than to the other. How will the speed of the air on the wide and narrow sides of the divider compare
Answer:
The answer is "The air will move faster on the narrow side".
Explanation:
The air on the top slows down in hypertensive. This is why light travels quicker on top. This results in air deflection downwards, required for its energy conservation to generate lift, and that is why air has to be moved quicker on the narrow side by the very same airflow per unit time as it departs.
what does it mean to have an acceleration of 8m/s^2
A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 5.0 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.
Required:
a. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.
b. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.
c. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.
Answer:
a) Nonconservative Work
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy
[tex]K_{f} = 15.614\,J[/tex]
Explanation:
The nonconservative work due to water resistance is defined by definition of work:
[tex]W_{disp} = F\cdot (y_{o}-y_{f})[/tex] (1)
Where:
[tex]W_{disp}[/tex] - Dissipate work, in joules.
[tex]F[/tex] - Resistance force, in newtons.
[tex]y_{o}[/tex] - Initial height, in meters.
[tex]y_{f}[/tex] - Final height, in meters.
The final gravitational potential energy ([tex]U_{f}[/tex]), in joules, is calculated by means of the definition of gravitational potential energy:
[tex]U_{f} = m\cdot g\cdot y_{f}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the rock, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
The final translational kinetic energy ([tex]K_{f}[/tex]), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:
[tex]m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}[/tex] (3)
[tex]K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}[/tex]
Lastly, the mechanical energy of the system ([tex]E[/tex]), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:
[tex]E = U_{f} + K_{f} + W_{disp}[/tex] (4)
Now we proceed to solve the exercise in each case:
a) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)[/tex]
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)[/tex]
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 0\,J[/tex], [tex]W_{disp} = 9\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J[/tex]
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0.50\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)[/tex]
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0.5\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)[/tex]
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 12.259\,J[/tex], [tex]W_{disp} = 6.5\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J[/tex]
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)[/tex]
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)[/tex]
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 24.518\,J[/tex], [tex]W_{disp} = 4\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J[/tex]
[tex]K_{f} = 15.614\,J[/tex]
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
a cleaner pushes a 4.50 kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of it's accerlation
Answer:
Explanation:
F = ma so filling in:
60.0 = 4.50a and
a = 13.3 m/s/s
You are cooking breakfast for yourself and a friend using a 1,140-W waffle iron and a 510-W coffeepot. Usually, you operate these appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period
Answer:
The cost is 297 cents.
Explanation:
Power of iron, P = 1140 W
Power of coffee pot, P' = 510 W
Voltage, V = 110 V
Time, t = 0.5 h each day
Cost = 12 cents per kWh
(a) Total energy
E = P x t + P' x t
E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60
E = 2052000 + 918000 = 2970000 J
1 kWh = 3.6 x 10^6 J
E = 0.825 kWh
For 30 days
E' = 0.825 x 30 = 24.75 kWh
So, the cost is
= 12 x 24.75 = 297 cents
A 70-kg astronaut (including spacesuit and equipment) is floating at rest a distance of 13 m from the spaceship when she runs out of oxygen and fuel to power her back to the spaceship. She removes her oxygen tank (3.0 kg) and flings it away from the ship at a speed of 15 m/s relative to the ship.
Required:
a. At what speed relative to the ship does she recoil toward the spaceship?
b. How long must she hold her breath before reaching the ship?
Answer:
Explanation:
mass of the astronaut including the spacesuit, [tex]M=30[/tex]
distance of astronaut from the spaceship, d = 13 m
mass of the oxygen tank, m = 3 kg
Speed of tank with respect to spaceship, [tex]v=15~m/s[/tex]
a)
Using the conservation of linear momentum:
total momentum before collision = total momentum after collision
[tex]M.u=m.v+(M-m)v'[/tex]
[tex]0=3\times 15+(70-15)\times v'[/tex]
[tex]v'=0.82~m/s[/tex]
b)
She mush hold her breath until she reaches the spaceship, i.e.
[tex]t=d/v'[/tex]
[tex]t=13/0.82[/tex]
[tex]t=15.89~s[/tex]
If you increase the charge on a parallel-plate capacitor from 3 mu or micro CC to 9 mu or micro CC and increase the plate separation from 1.8 mm to 5.4 mm, the energy stored in the capacitor changes by a factor of:__________
Explanation:
The energy stored in a capacitor is given by
[tex]U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}[/tex]
In the case of a parallel plate capacitor, the capacitance C is given by
[tex]C = \dfrac{\epsilon_0A}{d}[/tex]
so we can rewrite the expression for the energy as
[tex]U = \dfrac{Q^2d}{2\epsilon_0 A}[/tex]
Increasing the charge from [tex]3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C}[/tex] means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by
[tex]U' = \dfrac{Q'^2d'}{2\epsilon_0 A}[/tex]
[tex]\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}[/tex]
[tex]\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)[/tex]
[tex]\:\:\:\:\:\:= 27U[/tex]
As we can see, tripling both the charge and the separation distance result in the 27-fold increase in its stored energy U.
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall. What can you conclude? Under what assumptions? Give as much detail as you can.
Answer:
Ok, first, suppose that a given object is dropped or thrown down.
Then the acceleration of the object is the gravitational acceleration, given by:
a(t) = -9.8m/s^2
The velocity of the object can be integrated from that, it gives:
v(t) = (-9.8m/s^2)*t + V0
where V0 is the initial speed of the object, in the case that it is dropped, V0 is equal to 0m/s
The position equation can be found integrating again:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
Where p0 is the initial position.
Now let's see our problem.
We know that in 8 frames, the flowerpot falls 0.84 of the height of the window, which is 1.27m
Then the distance that it falls is:
D = 0.84*1.27m = 1.07m
Now we also know that the camera captures at 30 fps
Then we have the relation:
30 frames = 1 second
1 = (1 second)/(30 frames)
With this, we can rewrite:
8 frames = 8 frames* (1 second)/(30 frames) = 0.267 seconds.
So now we know the distance that the pot fall, and the time in which it did fall.
Remember that the position equation for a falling object, is:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t + p0
let's define p0 = 0m
Then the position equation of the pot is given by:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + V0*t
The distance that the pot would travel between t = 0s and t = t' is:
distance = p(t') - p(0s)
So, knowing that between t= 0s and t = 0.267s, the pot did travel -1.07m (the negative sign is because it traveled downwards), we can find the initial speed of the pot.
-1.07m = p(0.267s) - p(0s)
-1.07m = (1/2)*(-9.8m/s^2)*(0.267s)^2 + V0*0.267s
[-1.07m + (1/2)*(9.8m/s^2)*(0.267s)^2]/0.267s = V0 = -2.7 m/s
This means that the pot was thrown downwards with an initial speed of 2.7 m/s
So we can conclude that the pot did not fall down on its own, someone threw it intentionally downwards.
This is the only thing that we can conclude with the given information.
A current of 1 mA flows through a copper wire. How many electrons will pass a point in each second?
Answer:
A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown, assuming negligible air resistance
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
What is the projectile motion?An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.
The initial velocity is zero. Then the time taken to reach the ground is given as,
h = ut + 1/2at²
- 24 = 0×t + 1/2 (-9.8)t²
24 = 4.9t²
t² = 4.8979
t = 2.213 seconds
Then the horizontal speed is given as,
v = 18 / 2.213
v = 8.133 meters per second
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
Learn more about projectile motion:
https://brainly.com/question/29761109
#SPJ6
A plane flying horizontally at an altitude of 1 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
Now let's define the position of the radar as:
(0mi, 0mi)
Then we can write the position of the plane as:
(480mi/h*t, 1mi)
where t is time in hours.
Then we can write the distance equation as:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
Now we want to get:
the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
So first we want to find the value of t such that:
d(3) = 3mi
We will look at the positive value of t, because at this point the plane is increasing its distance to the station.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
The rate of change when the plane is 3 mi away from the station is:
d'(0.0059h)
remember that:
d'(t) = dd(t)/dt
We can write:
d(t) = h( g(t) )
such that:
h(x) = √x
g(t) = (480mi/h*t)^2 + (1mi)^2
then:
d'(t) = h'(g(t))*g'(t)
This is:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
the current through a wire is measured as the potential difference is varied what is the wire resistance
Answer:
Resistance, R = 0.02 Ohms
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
V = IR
Where;
V is the voltage or potential difference.
I is the current.
R is the resistance.
From the attachment, we would pick the following values on the graph of current against voltage;
Voltage, V = 0.5 V
Current = 25 A
To find resistance;
R = V/I
R = 0.5/2.5
Resistance, R = 0.02 Ohms
Note:
Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
Mộtchấtđiểmtăngtốctừ v0 =20m/sđạtđến v=80m/strênđườngthẳngvớigia
tốc a = 45 m/s2. Khoảng thời gian thay đổi tốc độ đó là bao nhiêu?
Answer:
Waktu, t = 1.33 detik
Explanation:
Mengingat data berikut;
Kecepatan awal, Vo = 20 m/s
Kecepatan akhir, Vi = 80 m/s
Percepatan, a = 45 m/s²
Untuk menemukan interval perubahan kecepatan;
Dalam latihan ini, Anda diminta menghitung waktu untuk perubahan kecepatan.
Oleh karena itu, kita akan menggunakan persamaan gerak pertama;
Vi = Vo + at
Mengganti ke dalam rumus, kita memiliki;
80 = 20 + 45t
80 - 20 = 45t
45t = 60
Waktu, t = 60/45
Waktu, t = 1.33 detik
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
Given the triangle shown below, what is the cosine of the angle 0 ?
triangle trig image 1
O Vx/V
O Vx/Vy
O Vy/Vx
O Vy/
Answer:
?
Explanation:
i am sorry but there is no triangle
Explanation: edmentum sample answer
what is the final velocity if you have an initial velocity of 5 m/s with an acceleration of 3 m/s^2 over a 4 second interval
Answer:
initial velocity (u)=5m/s
final velocity (v)=?
acceleration (a)=3m/s^2
time (t)=4s
now,
acceleration (a)=v-u/t
3=v-5/4
3×4=v-5
12=v-5
12+5=v
17=v
v=17
In the graph below, why does the graph stop increasing after 30 seconds?
A. The hydrogen gas is absorbing heat to undergo a phase change.
B. A catalyst needs to be added to increase the amount of hydrogen produced.
C. No more hydrogen can be produced because all of the reactants have become products at this point.
D. It has reached the maximum amount of product it can make at this temperature. The temperature would need to increase to produce more.
Answer:
The answer is "Option C".
Explanation:
It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.
hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.
A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.
How far away should a Cliff be from a source of sound to give an echo in 5.3 second?(given speed of sound at 0°c=331 m/s
Answer:
Total distance traveled by the sound wave = 2d . Thus we can conclude that the cliff should be at a distance of 877.15 metres from the source of sound to meet the required conditions
Explanation:
this is what i found on the web
Forced response is just the current or voltage that occures due to an input voltage or current.
a. True
b. False
Answer:
Option B
Explanation:
Generally
The forced response of a system is what the circuit does with the sources turned on, but with the initial conditions set to zero.
Therefore
Option B
Uranus and Neptune may have a compressed liquid water ocean beneath their atmospheres. What three pieces of evidence support this conclusion?
Answer:
Distance from sun, orbit and rotation. Presence of interior oceans, and elements that forms compressed water.
Explanation:
Both the planets are Jovian planets and have layers formed by ice such as that of Uranus does not have any surface. Such as the planet is only rotating fluids. While 80% of the mass of Neptune is made up of fluids or icy water and also consists of ammonia and methane.B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
True or False: The forces applied by our muscles on our bones are usually several times larger than the forces we exert on the outside world with our limbs.
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
An electron with an initial speed of 660,000 m/s is brought to rest by an electric field.
A) What was the potential difference that stopped the electron?
B) What was the initial kinetic energy of the electron, in electron volts?
Answer:
A) ΔV = 1.237 V
B) K.E = 1.237 eV
Explanation:
B)
The initial kinetic energy of the electron is given by the following formula:
[tex]K.E = \frac{1}{2}mv^2\\\\[/tex]
where,
K.E = Kinetic Energy of electron = ?
m = mass of elctron = 9.1 x 10⁻³¹ kg
v = speed of electron = 660000 m/s
Therefore,
[tex]K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2[/tex]
K.E = 1.98 x 10⁻¹⁹ J
K.E = (1.98 x 10⁻¹⁹ J)([tex]\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}[/tex])
K.E = 1.237 eV
A)
The energy applied by the potential difference must be equal to the kinetic energy of the electron, in order to stop it:
[tex]e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}[/tex]
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}[/tex]
ΔV = 1.237 V
plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the distance travelled and amount of force applied.
Answer:
Mass, M = 1000 kg
Speed, v = 90 km/h = 25 m/s
time, t = 6 sec.
Distance:
[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]
Force:
[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]
HELP NEEDED FAST (last cram sessions before finals)
BRAINLIEST!
Three resistors are connected in series across a 75-V potential difference. R, is 170 and R2 is 190. The potential difference across R3 is 21 V. Find the current in the circuit.
Explanation:
The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.
[tex]V_T = V_1 + V_2 +V_3[/tex]
From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as
[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]
[tex]I(170+190)=54\:\text{V}[/tex]
[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
