Answer:
A.Blackbody Radiation
Explanation:
Answer:
A. Blackbody radiation
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
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]
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]
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]
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]
A phosphodiester bond is used to:
A. Join glycerol to fatty acids
B. Join two nucleotides into a polynucleotide
C. Join two glucose molecules
D. Join two amino acids into a polypeptide
Answer:
A. Join glycerol to fatty acids
Explanation:
I majored in Physics.
What is the frequency of a wave with a Wavelength of 200 cm?
Explanation:
wavelength =200cm=2m
frequency is reciprocal of wavelength
f=1/2
f=0.5Hz or 1/s
The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.
Calculation of the frequency:Since the wavelength is 200 cm or 2m
Also we know that
frequency is reciprocal of wavelength
So,
f=1/2
f=0.5Hz
Hence, The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.
Learn more about frequency here: https://brainly.com/question/24470698
A certain electrical circuit contains a battery with three cells, wires, and a light bulb. Which of the following would cause the bulb to shine less brightly?
A. decrease the resistance of the circuit
B. increase the resistance of the circuit
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:
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.
learn more:
https://brainly.com/question/25955625
explain and derive the equation for capillary action in the phenomenon of surface tension
Answer:
Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.
Physical science deals with the ... of matter
2. Fossil fuel has been used more in the
existing world
Answer: fossil fuels are using more in the existing world because fossil fuels are of great importance because they can burned producing significant amount of energy prr unit mass the use of coal as a fuel prefates recorded history
Explanation:
Answer:
A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms, containing organic molecules originating in ancient photosynthesis[1] that release energy in combustion.[2] Such organisms and their resulting fossil fuels typically have an age of millions of years, and sometimes more than 650 million years.[3] Fossil fuels contain high percentages of carbon and include petroleum, coal, and natural gas.[4] Peat is also sometimes considered a fossil fuel.[5] Commonly used derivatives of fossil fuels include kerosene and propane. Fossil fuels range from volatile materials with low carbon-to-hydrogen ratios (like methane), to liquids (like petroleum), to nonvolatile materials composed of almost pure carbon, like anthracite coal. Methane can be found in hydrocarbon fields alone, associated with oil, or in the form of methane clathrates.
Coal, a fossil fuel
As of 2018, the world's main primary energy sources consisted of petroleum (34%), coal (27%), and natural gas (24%), amounting to an 85% share for fossil fuels in primary energy consumption in the world. Non-fossil sources included nuclear (4.4%), hydroelectric (6.8%), and other renewables (4.0%, including geothermal, solar, tidal, wind, wood, and waste).[6] The share of renewables (including traditional biomass) in the world's total final energy consumption was 18% in 2018.[7] Compared with 2017, world energy-consumption grew at a rate of 2.9%, almost double its 10-year average of 1.5% per year, and the fastest since 2010.[8]
Although fossil fuels are continually formed by natural processes, they are generally classified as non-renewable resources because they take millions of years to form and known viable reserves are being depleted much faster than new ones are generated.[9][10]
Most air pollution deaths are due to fossil fuel combustion products, it is estimated to cost over 3% of global GDP,[11] and fossil fuel phase-out would save 3.6 million lives each year.[12]
The use of fossil fuels raises serious environmental concerns. The burning of fossil fuels produces around 35 billion tonnes (35 gigatonnes) of carbon dioxide (CO2) per year.[13] It is estimated that natural processes can only absorb a small part of that amount, so there is a net increase of many billion tonnes of atmospheric carbon dioxide per year.[14] CO2 is a greenhouse gas that increases radiative forcing and contributes to global warming and ocean acidification. A global movement towards the generation of low-carbon renewable energy is underway to help reduce global greenhouse-gas emissions.
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 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} }
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]
This 200-kg horse ran the track at a speed of 5 m/s. What was the average kinetic energy?
Answer:
2500 JExplanation:
The average kinetic energy can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 200 \times {5}^{2} \\ = 100 \times 25[/tex]
We have the final answer is
2500 JHope this helps you
The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation. Enter your answer in accordance to the question statement
Answer:
1.04*10⁻⁵
Explanation:
light wave do showcase some behaviors whenever there is encounters with the end of the medium, some of the behaviors are - reflection, refraction, as well as diffraction. When visible light wave strikes a boundary that exist two different media, a portion of the energy will be transmitted into the new medium and some reflected.
Reflection of a light wave can be regarded as bouncing off of light wave from boundary. refraction on other hand is bending of the path of a light wave.
We were to calculate the reflectivity at the boundary,
reflectivity at the boundary can be calculated using the expression below
Reflectivity= (n₂ - n₁)² /(n₂ + n₁ ) ²
where
n₁= Indices of refraction at first grain= 1.545
n₂= Indices of refraction at second grain=
1.555
(1.555 - 1.545)² / (1.555 - 1.545)²
=(0.01)²/ (3.1)²
= 0.0001/ 9.61
= 1.04*10⁻⁵
Hence, the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation is 1.04*10⁻⁵
A stiuden goes for a bike ride ayt 20 meters
A convex mirror has a focal length of -12 cm. A lightbulb with
a diameter of 6.0 cm is placed 60.0 cm in front of the mirror.
Locate the image of the lightbulb. What is its diameter?
Answer:
PLEASE MARK AS BRAINLIEST!!
Explanation:
ANSWER IS IN THE IMG BELOW
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket, and the drive
Answer:
The minimum coefficient of static friction required, µ = 0.10
Note. The question is incomplete. The complete question is given below:
While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.
The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.
Explanation:
First, velocity in mph is converted to m/s
1 mph = 0.447 m/s
55 mph ≈ 24.6 m/s
The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²
Force that can be generated by the truck, F = ma
F = 8850kg * 1.07 m/s² = 9469.5 N
However, with the added mass of the log on it, the acceleration of the truck will become;
a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²
Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N
Normal reaction on the truck due to the weight of the log, R = mg
R = 929 kg * 9.8m/s² = 9104.2 N
Coefficient of static friction, µ = F/R
µ = 901.13/9104.2
µ = 0.098 ≈ 0.10
Therefore, the minimum static friction required is µ = 0.10
It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway (D-Hall), a distance of 47.0 meters. The second hallway (C-Hall) is filled with students, and she covers its 63.0 m length quickly. The final hallway (B- Hall) is empty, and Susie sprints its 76.0 m length. How fast does Susie need to go to make it to class on time (Hint: Calculate the total distance. Then calculate her total average speed rounded to the nearest tenths in meters/seconds.)?
Answer:
3.1 m/s
Explanation:
The total distance she has to run is the addition of the three lengths:
47 + 63 + 76 = 186 meters.
She needs to cover it one minute (60 seconds). Therefore her speed must be:
186 m / 60 s = 3.1 m/s
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.
Learn more about refraction: https://brainly.com/question/11155928
what are the importance of informal education?
Answer:
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Explanation:
Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Numerical Problems
A bus covers a distance of 600 metres in 30 seconds. Calculate the speed of
the bus.
Answer:
20 m/s.
Explanation:
From the question given above, the following data were obtained:
Distance travelled = 600 m
Time taken = 30 s
Speed =?
Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can easily calculate the speed of the bus as follow:
Distance travelled = 600 m
Time taken = 30 s
Speed =?
Speed = Distance /time
Speed = 600 / 30
Speed = 20 m/s
Therefore, the speed of the bus is 20 m/s.
I dont understand physics.
Answer:
same but from what i know is the newtons thingy
Explanation:
hope that helps!
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.
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.
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 would best help a student determine the net force acting on a rollercoaster car as it moves from one point on its track to another?
Answer:
you can show them a vid
Explanation: