Correct option is B 8 cm.
Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.
Parallel axes theorem gives:
[tex] {mr}^{2} = m( {k}^{2} + {a}^{2} ) \\ ⇒ {r}^{2} = {k}^{2} + a {}^{2} [/tex][tex]⇒k = \sqrt{ {10}^{2} - {6}^{2} } = 8cm.[/tex]Thus, option B is the correct answer.
For an object spinning around a central point, what will happen if its distance from the center is decreased?
A. Nothing will change.
B. Its acceleration will decrease.
C. Its acceleration will increase.
D. The centripetal force will decrease.
Answer:
Its acceleration will increase.
Explanation:
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity
This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves smaller radii and for lager radii acceleration will be less.
acceleration, a = v²/r
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
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Just before it strikes the ground, what is the watermelon's kinetic energy?
Answer:
Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.
Explanation:
This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some
It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t)=Ccos(ωt)+Ssin(ωt), where C, S, and ω are constants.
A) Using the general equation for x(t) given in the problem introduction, express the initial position of the block xinit in terms of C, S, and ω (Greek letter omega).
b) Find the value of S using the given condition that the initial velocity of the block is zero: v(0)=0.
c)What is the equation x(t) for the block? Express your answer in terms of t, ω, and xinit.
d)Find the equation for the block's position xnew(t) in the new coordinate system.
Express your answer in terms of L, xinit, ω (Greek letter omega), and t.
The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.
x = C cos wt + S sin wt
Where x is the displacement C and S are constants. W the angular velocity and t the time.
A) The initial position of the body occurs when the time is zero, t = 0
We substitute.
x = C cos 0 + S sin 0
[tex]x_{init}[/tex] = C
B) The velocity of the particle is defined.
[tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]
The initial velocity occurred for time zero t = 0
v = - S w
It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.
S = 0
C) The equation for the block remains.
x (t) = [tex]x_{init} \ cos \ wt[/tex]
D) In some cases it is measured with respect to another reference system, the most common are:
For maximum compression it is the zero of the system. The maximum extension is the zero of the system.
In these cases, the change that must be made is
x = [tex]L - x_{min}[/tex] t
we substitute
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
L = [tex]x_{init}[/tex] (1 + cos wt)
In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
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Please help me as quick as possible!!!!!!! Please, please please!!!!!!
Answer:
23
Explanation:
3x-4
A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction?
The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
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A truck moves 60 km West, and then 80 km North, and then
travels in a straight line back to its starting point. The distance
travelled by the truck is ____km and its displacement is _____km
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Distance travelled by the truck is ~
[tex] \boxed{240 \: \: km}[/tex]And it's displacement is ~
[tex] \boxed{0 \: \: km}[/tex][tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]
See the diagram in attachment for reference ~
Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.
The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~
According to Pythagoras theorem ~
hypotenuse² = sum of squares of other two legs
that is ~
[tex]h {}^{2} = 60 {}^{2} + 80 {}^{2} [/tex][tex] {h}^{2} = 3600 + 640 0[/tex][tex]h {}^{2} = 10000[/tex][tex]h = \sqrt{10000} [/tex][tex]h = \sqrt{100 \times 100}{}[/tex][tex]h = 100 \: \: km[/tex]So, the distance between the point A and O is 100 km
Now, The total distance is equal to the distance covered through actual path that is ~
60 km + 80 km + 100 km 240 kmAnd displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.
How much energy is consumed by a 12 W night light left on for 10 hr?
Answer:
Energy consumed is 0.00033 Joules.
Explanation:
the formula of Energy is:
Energy = power/ time.
A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.
Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first
Answer:
The box arrives first.
Explanation:
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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
What is Friction?Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.
There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.
Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
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Peregrine falcons, which can dive at 200 mph (90 m/s), grab prey birds from the air. The impact usually kills the prey. Suppose a 480 g falcon diving at 75 m/s strikes a 240 g pigeon, grabbing it in her talons. We can assume that the slow-flying pigeon is stationary. The collision between the birds lasts 15 ms.
Answer:
What is the average force of collision?
Explanation:
The velocity of the combined mass after impact is found by conservation of momentum
0.480(75) + 0.240(0) = (0.480 + 0.240)v
v = 50 m/s
An impulse results in a change of momentum
FΔt = mΔv
F = mΔv/Δt
for the pigeon
F = 0.240(50 - 0)/0.015
F = 800 N
for the falcon
F = 0.480(50 - 75)/0.015
F = -800 N
The final speed(v) of the Peregrine falcons and pigeon is 50 m/s
The objective of this question is to determine the final speed(v) of the Peregrine falcons and pigeon
From the information given:
the mass of the falcon m_f = 480 gthe speed of the falcon v_f = 75 m/sthe mass of the pigeon m_p = 240 gthe collision time = 15 ms = 0.015 sAccording to the conservation of momentum, we can say that the totality of momentum before and after the collision is the same. As such;
[tex]\mathbf{m_fv_f = (m_f+m_p) v}[/tex]
[tex]\mathbf{480 g \times 75 m/s = (480 + 240) \ g \times v}[/tex]
[tex]\mathbf{v = \dfrac{480 g \times 75 m/s}{(480 + 240) \ g}}[/tex]
[tex]\mathbf{v = \dfrac{36000 \ m/s}{ 720}}[/tex]
v = 50 m/s
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A friend has suggested that you go swimming in a pool having water of temperature 350 K. What would this temperature be on the Fahrenheit scale?
109°F
123°F
170°F
202°F
This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.
The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.
the relation between Kelvin scale and Fahrenheit scale is given by:
(F - 32)/180 = (K - 273)/100
F - 32 = (350 - 273)(9/5)
F = 32 + (350 - 273)(9/5)
F = 170
Hence, this temperature would be 170° F on the Fahrenheit scale.
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A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.
The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
The given parameters;
length of the cubical box, L = 0.368number of moles of the gas, n = 1.98temperature, T = 298 Kmass of the gas, m = 3.35 x 10⁻²⁶ kgThe average kinetic energy of the gas molecules is calculated as follows;
[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]
The average speed of the gas molecules is calculated as follows;
[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]
The time of collision of the gas molecules with the walls of the container is calculated as follows;
[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]
The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;
[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]
Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
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Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.
We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."
Answers:
Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex] Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]
From the question we are told
The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s
A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube
Therefore, the mass flow rate of CCI_4 at point A
= [tex]5.86*10^{-13} kg/s[/tex]
B) From Fick's law
[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]
Then,
[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]
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Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that the red car has ____ mass as the blue car.
Answer:
equal
Explanation:
help me for a physics project please
Mister Brainly Please Help Me
Write 10 Information's About Sound
-cant travel through space since there's no molecules to travel through
-sound travels 4.3 times faster in water than air
-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules
-different types of sound like audible, inaudible, infrasonic, ultrasonic,
-sounds waves are either longitudinal, mechanical and pressure waves
-sound travels at 767 miles per hour
Contrast the behavior of a water wave that travel by a stone barrier to a sound wave that travels through a door
calvin carter
Explanation:
here the file has everything
electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection? why
Answer:
Explanation:
The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity. The deflection is also inversely proportional to its mass. So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.
What is the formula of [tex]\underline{force}[/tex]in physics?
The formula of force is F = ma.
Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton (N). Force is a vector quantity
Types of force
Friction forcePushPullUpthrustTensional force.
⇒ The formula of force is
F = ma................ Equation 1
⇒ Where:
F = forcem = mass of the bodya = accelerationHence, the formula of force is F = ma.
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Before looking at the formula, Lets understand what is force?
Force:- Force can be defined as the cause of motion of an object. Force can change the shape, direction and velocity of an object.It's SI unit is NewtonThe formula used to find force is:
[tex] \sf \nrightarrow \: F=m×a[/tex]
Where,
F is forceM is mass of the bodyA is accelerationEquation in words:- Force applied on an object is directly proportional or equal to the product of the mass of object and the acceleration of object.
I NEED THE ANSWER ASAPP
Answer:
Explanation:
a) The spring force will equal the weight.
b) If up is positive
kx - mg = 0
mg = kx kx = 25 N
c) m = kx/g = 25/10 = 2.5 kg
Convert 6 picoseconds into seconds.
Answer:
6e-12
Explanation:
divide the time value by 1e+12
Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 13 m/s due north. Plane 2 taxies with a speed of 8.5 m/s in a direction 20 ∘ north of west.
Part A
What is the magnitude of the velocity of plane 1 relative to plane 2?
Part B
What is the direction of the velocity of plane 1 relative to plane 2?
Part C
What are the magnitude of the velocity of plane 2 relative to plane 1?
Answer:
Explanation:
Plane 2 is moving north at
8.5sin20 = 2.9 m/s
Plane 2 is moving west at
8.5cos20 = 8.0 m/s
Part A
v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s
Part B
θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E
Part C
13 m/s 52° S of W
relative velocity magnitude is independent of reference frame
d what is
7 A rocket of mass 10000 kg uses 5.0kg of fuel and oxygen
to produce exhaust gases ejected at 5000 m/s. Calculate the
increase in its velocity
Answer:
Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.
Explanation:
Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.
Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.
The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.
Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].
The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].
The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:
[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].
The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:
[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:
[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:
[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].
[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.
[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].
[tex]9995\, \Delta v = 25000[/tex].
[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].
Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.
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Answer:
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A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
Swim swim swim swim swim swim swim swim swim swim swim swim.
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.
The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.
Answer:
Explanation:
Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.
In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.
That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.
Which properties make a metal a good material to use for electrial wires
Answer:
Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.
Lab report on velocity of sound