Answer:
C. A. B.
Explanation:
Credit to Lainey in the comments for the answer!
Answer:
1] C
2] A
3] B
Explanation:
if u want to play among us have fun the code is UQEZFF
Answer:
ight bet
Explanation:
A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of the circle so that the cord remains taut and the ball’s path remains circular.
Answer:
The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.
Explanation:
By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:
[tex]K =U_{g}[/tex] (1)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]U_{g}[/tex] - Gravitational potential energy, measured in joules.
By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h[/tex]
[tex]v = \sqrt{2\cdot g\cdot h}[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]h[/tex] - Maximum height of the ball, measured in meters.
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]h = 5\,m[/tex], then the initial speed of the ball is:
[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]
[tex]v\approx 9.903\,\frac{m}{s}[/tex]
The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.
During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approximately 103 cm/s. The blood volume traverses the aortic arch, exiting with the same speed but opposite direction. Assume the mass density of the blood is 1060 kg/m3 blood, the aortic arch remains stationary, and that the heart rate is 59 bpm. What is the average force exerted by the blood on the wall of the aorta.
Answer:
The value [tex]F = 0.1396 \ N [/tex]
Explanation:
From the question we are told that
The volume blood ejected is [tex]V = 65 \ cm^3 = 65*10^{-6} \ m^3[/tex]
The velocity of the blood ejected is [tex]v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s[/tex]
The density of blood is [tex]\rho = 1060 \ kg/m^3[/tex]
The heart beat is [tex]R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps[/tex]
The average force exerted by the blood on the wall of the aorta is mathematically represented as
[tex]F = 2 * \rho * V * R * v[/tex]
=> [tex]F = 2 * 1060 * 65*10^{-6} * 0.9833 * 1.03[/tex]
=> [tex]F = 0.1396 \ N [/tex]
In the Sun, fusion reactions create helium nuclei. To form each helium
nucleus, four hydrogen nuclei fuse. The four hydrogen nuclei have a greater
total mass than the newly formed helium nucleus. Which statement explains
this difference in mass?
O A. Some of the mass burned and was transformed into gases.
O B. Mass was destroyed and disappeared.
O c. Some of the mass of the four hydrogen nuclei was converted into
energy
D. Some of the mass was transformed into protons.
Answer:
C. Some of the mass of the four hydrogen nuclei was converted into
energy
Explanation:
a p e x , just took the quiz
Some of the mass of the four hydrogen nuclei was converted into energy.
When four hydrogen nuclei fuse to form one helium nuclei, mass defect, some mass is converted into energy, that is the reason of energy of the sun.
What is fusion?"When two nuclei form a big nuclei, the phenomenon is known as fusion."
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The direction of electrical force is based on the fact that like charges will
yes
Explanation:
because hindi ko tlaga ala
Answer:
the answer is repel
Explanation:
How can a particle's position determine the
amount of potential energy available to a
system?
Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.
The potential energy of a particle is the energy possessed by the particle by virtue of its position. This potential energy is form of mechanical energy and it is expressed mathematically as the product of the mass of the particle, acceleration due to gravity and the particles displacement or position.
P.E = mgh
where;
P.E is the potential energy of the particle
m is the mass of the particle
g is the acceleration due to gravity
h is the position of the particle.
Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.
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A hammer strikes a nail with a 10N force for .01 seconds. Calculate the impulse of the hammer
Answer:
Impulse = 0.1 Kgm/s
Explanation:
Given the following data;
Force, F = 10N
Time, t = 0.01 seconds
To find impulse
An impulse can be defined as the net force acting an object for a very short period of time.
Mathematically, impulse is given by the formula;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 0.01
Impulse = 0.1 kgm/s
Therefore, the impulse of the hammer is 0.1 kilogram meter per seconds.
The impulse of the hammer is 0.1 Ns.
To solve the given problem we need to use the formula for calculating impulse.
Impulse: This can be defined as the product of force and time on a body.
The formula of impulse is given below.
I = Ft....................... Equation 1
Where :
I = Impulse of the harmerF = Force on the nailt = timeFrom the question,
Given:
F = 10 Nt = 0.01 sSubstitute these values into equation 1
I = 10(0.01)I = 0.1 NsHence, The impulse of the hammer is 0.1 Ns.
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Resonance frequency
Answer:
resonance frequency is the frequency when capacitive reactance and inductive reactance become equal and opposite to each other and all impedence is given by resistance.
Explanation:
f=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]
Physical science deals with the ... of matter
In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m and 0.6 m, respectively. The wavelength is 630 nm. Suppose that the radius of the aperture can be increased from 0.5 mm, determine: (a) The first two radii when the center intensity at the reception screen is maximum. (b) The first two radii when the center intensity is minimum.
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
a consequence of "yoyo" dieting is that
Answer:
srry i ll get back to with an answer
Explanation:
At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2.
Calculate the distance travelled by car.
Answer:
62.5m
Explanation:
Given parameters:
Initial velocity = 0m/s
Time = 5s
Acceleration = 5m/s²
Unknown:
Distance traveled = ?
Solution:
To solve this problem, we use the motion equation given below:
S = ut + [tex]\frac{1}{2}[/tex] at²
S is the distance traveled
u is the initial velocity
a is the acceleration
t is the time taken
Now, insert the parameters and solve;
S =( 0 x 5) +( [tex]\frac{1}{2}[/tex] x 5 x 5²) = 62.5m
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
that you must drop into the bucket. Where should the bucket be when you drop the mass (so it is moving
purely vertically when it lands in the bucket) if your goal is to:
(a) Maximize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
(b) Minimize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t = [tex]\sqrt{2y_{o}/g}[/tex]
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w = [tex]\sqrt{k/m}[/tex]
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = - [tex]\sqrt{k/3}[/tex] [tex]\sqrt{2y_{o}/g}[/tex]
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Which lists three organic biological molecules?
O carbohydrates, salts, metals
O salts, proteins, minerals,
O proteins, lipids, carbohydrates
O lipids, metals, minerals
Answer:
B
Explanation:
I'm learning it in science.
Answer:
its not b i just took the test and b was wrong
Explanation:
What kind of electricity does turning wheel generates? Please help!
Answer: Kinetic Energy to Electrical.
Explanation: The magnet is rotated as a result of the spinning wheels, and this results in a powerful stream of electrons, therefore converting kinetic to electrical.
A cross-country skier travels 40 meters north in 2 minutes. What is the skiers velocity?
Answer:
v = 0.33 [m/s]
Explanation:
We must remember that speed is defined as the relationship between the displacement in a given time. In this way, we can propose the following equation.
[tex]v=d/t[/tex]
where:
v = velocity [m/s]
d = displacement = 40 [m]
t = 2 [min] = 120 [s]
Now replacing we have:
[tex]v=40/120\\v=0.33[m/s][/tex]
An automobile travels to the right, with the center Aof the wheel moving at a constant speed of 48 mi/h. If the diameter of a wheel is 22in.,determine the velocities of points B, C, D, and E on the rim of the wheelassuming there is no slipping. [Hint: at any given instant the velocity of point C, which is in contact with the ground must be zero]. Please introduce axes and give your answers in vector form.
Answer:
Explanation:
From the question we are told that
Constant speed =48mile/hr
Diameter of a wheel = 22inch therefore [tex]r=\frac{22}{2} =>11[/tex]
Generally Convert from mile/hr to inches/sec
The length in inches is equal to the miles multiplied by 63,360.
an hour is 3600seconds
[tex]\frac{48*63360}{3600}[/tex]
48miles/h = 849.8 inch/sec
[tex]V_a =844.8 inch/sec[/tex]
therefore
[tex]\omega= \frac{v}{r}[/tex]
[tex]\omega= \frac{844,8}{11}[/tex] =>[tex]76.8 sec ^-^1[/tex]
a)Considering the velocity of Vb in inches per seconds
Generally the formula is stated as
[tex]V_b= V_a + V_b_/_a[/tex]
[tex]V_b = 844.8 + \omega r[/tex]
[tex]V_b= 1639.6in/s[/tex]
b)Considering the velocity of Vc in inches per seconds
Since the tire doesn't slip as earlier stated in the question
Therefore [tex]V_c = 844.8 -w(r) =0[/tex]
c)Considering the velocity of Ve in inches per seconds
Generally the formula is stated as
[tex]V_e=V_a + V_e_/_a[/tex]
[tex]V_e = 844.8 \uparrow \theta -844.8 \uparrow = 844.8\sqrt{2}[/tex]
Expressing result with vector
[tex]V_e =844.54 + 20.85j[/tex]
d)Considering the velocity of Vd in inches per seconds
Generally the formula is stated as
[tex]V_d= V_a + V_d_/_a[/tex]
Mathematically
[tex]V_d =844.8 \uparrow +( 844.8\frac{\sqrt{3} }{2} \uparrow + \frac{844.8}{2} \uparrow)[/tex]
[tex]V_d= (1576.4 \uparrow + 422.4\uparrow)[/tex]
[tex]V_d= 1632.028in/s[/tex]
A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?
Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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What kind of energy is produce when sun reaches solar panel?
Answer:
Radient to ElEcTrIcAAl
Explanation:
The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.
A rightward force of 4.0 N is exerted upon an object for a distance of 3.0 meters.
What is the work done on the object?
Answer:
W = 12 J
Explanation:
Given that,
Force, F = 4 N
The object moves in rightward direction for a distance of 3 m.
Work done on the object is given by :
[tex]W=F\times d\\\\=4\ N\times 3\ m\\\\=12\ J[/tex]
So, the work done on the object is 12 J.
ANSWER ALL QUESTIONS and show your work ON THE ATTACHMENT AWARD 50 pts if you don’t know the answer to all of them I have posted them individually so I can still mark you brainlessly (however you spell the darn thing)
Answer:
the velocity is 210 - hq squared which equals to 14
The motion of a nightingale's wingtips can be modeled as simple harmonic motion. In one study, the tips of a bird's wings were found to move up and down with an amplitude of 8.0 cm and a period of 0.80 s.
Part A: What is the wingtips' maximum speed?
Part B: What is the wingtips' maximum acceleration?
Answer:
PART A: Maximum speed = 0.314 m/s
PART B: Maximum acceleration = 1.23 m/s²
Explanation:
A simple Harmonic motion is a repetitive motion through an equilibrium point.
Amplitude = 8.0cm = 8/100 = 0.08m (highest displacement)
period (T) = 0.80s
A) maximum speed [tex](V_{max)[/tex]
[tex]V_{max} = 2\pi fA\\where:\\A = Amplitude = 0.08m\\f = frequency = \frac{1}{period(T)} = \frac{1}{0.8} = 1.25 Hz\\\therefore 2\pi fA = 2\pi \times 1.25 \times 0.08\\= 0.314\ m/s[/tex]
B) maximum acceleration [tex](a_{max})[/tex]
[tex]a_{max} = (2\pi f)^2A\\where:\\f = 1.25Hz\\A = 0.08m\\a_{max} = (2\pi \times 1.25)^2 \times 0.08\\= 1.23\ m/s^2[/tex]
help please Derive an equation
Ta=1.44T1/2
Explanation:
To derive an equation you must indicate the variable you want to solve for.
Here we have tension of an object A and Tension 1.
Two variables or unknown are given hence we cannot derive any other equations.
types of magnetic resonance image
Answer: ur mom
Explanation:
A stiuden goes for a bike ride ayt 20 meters
A rifle is aimed at a target 40m away. The bullet hits the target 2.2cm below the horizontal plane (ignore wind and rotational effects).
(a) What is the time of flight for bullet?
(b) What is the muzzle velocity (velocity of bullet when it leaves the rifle)?
Answer:
The value is [tex]u_x = 597 \ m/s[/tex]
Explanation:
From the question we are told that
The distance of the target from the riffle is [tex]d = 40 \ m[/tex]
The height at which the bullet hit the target is [tex]y = 2.2 \ cm = 0.022 \ m[/tex]
Considering the vertical motion
Generally from kinematic equations we have that
[tex]y = u_y t + \frac{1}{2} * gt^2[/tex]
At the initial stage the velocity is zero i.e [tex]u_y = 0 \ m/s[/tex]
=> [tex]0.022 = 0 * t + \frac{1}{2} * 9.8 t^2[/tex]
=> [tex]t = 0.067 \ s[/tex]
Generally the velocity of the bullet when it leaves the riffle is mathematically represented as
[tex]u_x = \frac{ d}{t}[/tex]
=> [tex]u_x = \frac{40 }{ 0.067 }[/tex]
=> [tex]u_x = 597 \ m/s[/tex]
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 59.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.852 m/s2. Calculate her mass.
Answer:
The value is [tex]m = 69.24 \ kg[/tex]
Explanation:
From the question we are told that
The value of the external force is [tex]F = 59.0 \ N[/tex]
The magnitude of the astronaut's acceleration is [tex]a = 0.852 \ m/s[/tex]
Generally Newton's Second Law of Motion from the mass of the astronauts is mathematically represented as
[tex]m = \frac{F}{a}[/tex]
=> [tex]m = \frac{59 }{0.852 }[/tex]
=> [tex]m = 69.24 \ kg[/tex]
Which wave has a greater frequency
Answer:
A I think
Explanation:
because what is the most frequency a because it has more frequency I think I'm not that sure
Pelican hunts fish in the water as th pelican the waters surface why must it aim for the fish in a slightly different place than where the fish appears to be located
The Pelican must aim differently than where the fish appears to be because light is refracted from air to water.
Refraction is the change in the direction of a wave at the interface between two media. The effects of refraction are easily seen when light is travelling from air to water hence objects are not really where they appear to be when viewed above the water surface.
Hence, the Pelican must aim differently than where the fish appears to be because light is refracted from air to water.
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Consider a rigid 3-mass system ( with origin at the leftmost mass 1 kg) which can rotate about an axis perpendicular to the system. The mass are separated by rods of length 5m, so that the entire length is 10m. Find the x-coordinate of the center of the mass for the three-mass system with respect to the origin. Treat mass as particles. Answer in unit of m.
now consider a rotation axis perpendicular to the system and passing through the point Xo at distance 3.8 m from the leftmost mass 1kg. find the moment of inertia of the 3-mass system about the new axis. Answer in unit of kg.m^2
Answer:
1) x_{cm} = 5 m , 2) I = 168.32 kg m²
Explanation:
1) An important concept of center of mass is
[tex]x_{cm} = \frac{1}{M} \sum x_{i} m_{i}[/tex]
where M is the total mass of the system
Let's apply this equation to our case, suppose that all masses are equal and are worth 1 kg
[tex]x_{cm}[/tex] = ⅓ (1 0 + 1 5 + 1 10)
x_{cm} = 5 m
2) In this para indicates that there is an axis of rotation at the point xo = 3.8 m and they ask to calculate the moment inertia.
Let's use the parallel axes theorem
I = I_{cm} + M D
where I_{cm} is the moment of inertia with respect to the center of mass, D the distance between the two axes of rotation and M the total mass of the system
Let's look for the moment of inertia of the center of mass
[tex]I_{cm}[/tex] = 1 0 + 1 5² + 1 10²
I_{cm} = 125 kg m²
the total moment of inertia is
I = 125 + 3 3.8²
I = 168.32 kg m²
The moment of inertia of the 3-mass system about the new axis is 54.32 kgm/s².
We have three masses each of mass = 1kg such that they are in line with mass m at origin, m at 5m and m at 10m
(a) The center of mass:
[tex]X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X =\frac{ 1*0+1*5+1*10}{1+1+1}\\ \\ X = 5m[/tex]
Hence the center of mass of the system is at x = 5m.
(b) The moment of inertia about the axis passing through x = 3.8m
from the parallel axis theorem:
[tex]I = I_{cm} + Md^2[/tex]
where, [tex]I_{cm}[/tex] is the moment of inertial along an axis passing through the center of mass of the system, M is the total mass of the system and d is the distance of the given axis from center of mass.
M = 3kg
d = 5 - 3.8 = 1.2m
[tex]I_{cm}=1*5^2+1*0+1*5^2\\\\ I_{cm}=50 kgm/s^2[/tex]
Md² = 3×(1.2)²
Md² = 4.32 kgm/s²
I = 50 + 4.32
I = 54.32 kgm/s² is the moment of inertial about the given axis.
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