Answer:
Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1
Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2
(Xcm , Ycm) = (1 , 1/2)
Using definition of center of mass
A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?
Answer:
201.5537 mph
Explanation:
Given the following data;
Speed = 90.1 m/s
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
Speed = distance/time
To convert this value into miles per hour;
Conversion;
1 meter = 0.000621 mile
90.1 meters = 90.1 * 0.000621 = 0.05595 miles
1 metre per second = 2.237 miles per hour
90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour
90.1 m/s = 201.5537 mph
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]
So, the new work is more than 130 J.
Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off what
Answer:
Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)
Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.
Then the correct answer is a nickel target.
"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"
The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Explanation:
The period T of a simple pendulum is given by
[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]
Doubling the length of the pendulum gives us a new period T'
[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]
two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations
Answer:
The height is the same
Explanation:
Because they were at the same height but they fell at different velocities
1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?
N = 52.0 N
Explanation:
Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]
[tex]m=6.00\:\text{kg}[/tex]
[tex]N = \text{normal force}[/tex]
The net force [tex]F_{net}[/tex] is given by
[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]
Solving for N, we get
[tex]N = mg - F_a\sin 20[/tex]
[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]
[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]
The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?
Answer:
g₂ = 11 m/s²
Explanation:
The value of free-fall acceleration on the surface of a planet is given by the following formula:
[tex]g = \frac{Gm}{r^2}[/tex]
where,
g = free-fall acceleration
G = Universal Gravitational Constant
m = mass of the planet
r = radius of planet
FOR PLANET 1:
[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)
FOR PLANET 2:
[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]
using equation (1):
[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]
g₂ = 11 m/s²
An aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft dears the building by 12 m, how far from the building does the aircraft touch down on the runway
The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.
For every 12 meters it travels it drops 1 m.
Divide the height by 12 to find the distance it travels:
57 / 12 = 4.75
It touches down 4.75 meters from the building.
The building is 684 meters away from the aircraft touching down on the runway.
What are trigonometric functions?A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.
As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,
the total height of the aircraft when it clears the building = 45 +12
the total height of the aircraft when it clears the building is 57 meters
It is given that the Glide ratio is 12:1,
The distance of the building from touch down on the runway = 12 ×57
The distance of the building from the touch-down on the runway is 684 meters.
Thus, the building is 684 meters away from the aircraft touching down on the runway.
Learn more about the trigonometric functions here,
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Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?
Answer:
Converging (convex) lens.
Explanation:
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.
Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.
Mary needs to row her boat across a 160 m-wide river that is flowing to the east at a speed of 1.5 m/s. Mary can row with a speed of 3.6 m/s. If Mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore? What is her speed with respect to the shore?
Answer: 66.67 m, 44.44 s
Explanation:
Given
Velocity of flow is [tex]u=1.5\ m/s[/tex]
Mary can row with speed [tex]v=3.6\ m/s[/tex]
Width of the river [tex]y=160\ m[/tex]
Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction
time taken to cross river
[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]
Flow will drift Mary by
[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]
Velocity w.r.t shore is
[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm
Answer:
0.113 J
Explanation:
Applying,
w = ke²/2................. Equation 1
Where w = workdone in stretching the spring, k = spring constant, e = extension
make k the subject of the equation
k = 2w/e²................ Equation 2
From the question,
Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m
Substitute these values into equation 2
k = (2×3)/0.17²
k = 6/0.17
k = 35.29 N/m
(a) if the spring from 37 cm to 45 cm,
Then,
w = ke²/2
Given: e = 45-37 = 8 cm = 0.08
w = 35.29(0.08²)/2
w = 0.113 J
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
explain why sound wave travel faster in liquid than gas
Answer:
Because gas contains free molecules but not liquid.
Please mark as brainliast
You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:
Answer:
18
Explanation:
I'm pretty sure I got it right
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.
Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?
(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is
W = - (1/2 kx ² - 1/2 k (0.0400 m)²)
(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so
W = ∆K = 0 - 1/2 m (0.835 m/s)²
Solve for x :
- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> x ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
W = - 1/2 k (0.0400 m)²
If v is the glider's maximum speed, then by the work-energy theorem,
W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²
Solve for v :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²
==> v ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s
From the given information;
the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 mUsing energy conservation E, the amplitude of the motion can be calculated by using the formula:
[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]
[tex]\mathbf{E =0.194 \ J}[/tex]
Similarly, we know that:
[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]
Making amplitude A the subject, we have:
[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]
[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]
[tex]\mathbf{A =0.049 \ cm}[/tex]
Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:
[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]
[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]
The angular frequency of the oscillation can be computed by using the expression:
[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]
[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]
ω = 29.02 rad/s
Learn more about energy conservation here:
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A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic field for that wave at P
Answer:
[tex]6.63\times 10^8\ N/C[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 2.21
We need to find the magnitude of the electric field. Let it is E. So,
[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]
Put all the values,
[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]
So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].
The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.
Answer:
Look at explanation
Explanation:
I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.
The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.56 m/s²
Explanation:
Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.
The total time (time of flight) of an object is given by:
T = 2usinθ / g
where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity
Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.
T = 2usinθ / g
Substituting:
24.4 = 2(19)(sin90)/g
g = 2(19)(sin90) / 24.4
g = 1.56 m/s²
No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?
Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito
Answer:
fair palybtgshsisuehdh
Images formed by a convex mirror are always
Answer:
Images formed by a convex mirror are always virtual
Explanation:
A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.
OAmalOHopeO
A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column
Answer:
665.25 Pa
Explanation:
Given data :
Weight of the disk, w = 45 N
Diameter, d = 30 cm
= 0.30 m
Therefore, radius of the disk,
[tex]$r=\frac{d}{2}$[/tex]
[tex]$r=\frac{0.30}{2}$[/tex]
= 0.15 m
Now, area of the cylindrical disk,
[tex]$A=\pi r^2$[/tex]
[tex]$A=3.14 \times (0.15)^2$[/tex]
[tex]$=0.07065 \ m^2$[/tex]
∴ The gauge pressure at the top of the oil column is :
[tex]$p=\frac{w}{A}$[/tex]
[tex]$p=\frac{47}{0.07065}$[/tex]
= 665.25 Pa
Therefore, the gauge pressure is 665.25 Pa.
The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa
The pressure is defined by the relationship between perpendicular force and area.
[tex]P = \frac{F}{A}[/tex]
where P is pressure, F is force, and A is area.
They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.
The area is:
A = π r² = [tex]\pi \frac{d^2}{4}[/tex]
In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.
B-W = 0
B = W
The force applied to the liquid is the weights of the cylinder. Let's replace.
[tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]
Let's calculate.
[tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²
P = 636.6 Pa
In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa.
Learn more about pressure here: brainly.com/question/17467912
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb
Answer: [tex]29.85\ W/m^2[/tex]
Explanation:
Given
Power [tex]P=60\ W[/tex]
Distance from the light source [tex]r=0.4\ m[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Inserting values
[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]
Answer:
29.85 W/ m^2
Explanation:
The value of mass remains constant but weight changes place to place why
Explanation:
No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).
Explanation:
Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
