The radiative zone and core of the Sun are thought by scientists to revolve more like solid bodies than liquid ones. From the convective zone outward, the Sun's outer regions rotate at various rates that change with latitude.
What is latitudes?It is calculated using 180 fictitious lines that are drawn in circles east-west of the equator. Parallels are the names for these lines.Because the Sun is made of gaseous plasma, its rotation varies with latitude. At the equator, the rate of rotation is seen to be the quickest, and it slows down as latitude rises.Since it is made of gas and plasma, the rotational velocities of the gasses and plasma vary depending on where they are on the sun. Every 25 days, the gas and plasma close to the equator of the sun rotate around its axis.Polaris will appear lower in the sky as you move further south, and the sun will set at nighttime farther and more quickly.To learn more about latitudes refer to:
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A 0. 0850-kg arrow is fired horizontally. If the bowstring exerts an average force of 89. 0 n on the arrow over a distance of 0. 782 m, with what speed does the arrow leave the bow?.
40.47 m/s is the speed at which the arrow departs the bow.
Does velocity not depend on gravity?
An object's horizontal velocity is constant for small displacements and operates independently of the vertical force of gravity. The horizontal and vertical halves of an object that is projected at an angle to gravity can also be separated. The force of gravity has no impact on the horizontal velocity component either.
The arrow will start moving and accelerating as soon as it is let go from the bow. The arrow obeys the second law of motion, according to Newton. The method
m = mass of the arrow (kg) = 0.0850 kg
a = acceleration of the arrow (m/s²)
F = force (N) = 89.0 N
F=m*a
a = 1,047.059 m/s²
When an object accelerates, it will travel in a non-uniform manner. The method
[tex]v^{2} = v_{0} ^{2} +2*a*d[/tex]
= 40.47 m/s
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A driver spotted a police car and he braked from a speed of 100km/h to a speed of 80km/h during a displacement of 88m, at a constant acceleration (a) What is that acceleration. (b) How much time is required for the given decrease in speed
Answer:
a = − 1.579 [tex]\frac{m}{s^{2} }[/tex]
b. t = 3.52s
We have velocity in km/hr and displacement in m. Since we have to make a unit conversion, might as well do it so it will yield the answer in the normal (SI) units for acceleration.
u = 100 x [tex]\frac{5}{18}[/tex] = 27.78 [tex]\frac{m}{s}[/tex]
v = 80 x [tex]\frac{5}{18}[/tex] = 22.22 [tex]\frac{m}{s}[/tex]
Using the formula of motion [tex]v^{2} - u^{2}[/tex] = 2*a*d
[tex](22.22)^{2}[/tex] - [tex](27.78)^{2} \\[/tex] = 2 * a * 88
a = -1.579 [tex]\frac{m}{s^{2} }[/tex] = acceleration
Since the acceleration was constant, we can use the average velocity, and the distance to determine the time.
time = [tex]\frac{displacement}{speed}[/tex]
t = [tex]\frac{88}{speed}[/tex]
speed = [tex]\frac{1}{2}[/tex] * (u+v) = [tex]\frac{1}{2}[/tex] * (22.22 + 27.78) = 25[tex]\frac{m}{s}[/tex]
t = [tex]\frac{88}{25}[/tex] = 3.52 s
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The acceleration of the car is equal to -1.58 m/s. The time required for the given decrease in speed is equal to 3.52 s.
What are equations of motion?The equations of motion can be utilized to describe the relationship between the acceleration, velocity, time, and displacement of a moving object.
The equations of motions can be mathematically expressed as shown below:
[tex]v = u + at\\S= ut +\frac{1}{2}at^2\\ v^2-u^2 = 2aS[/tex]
Given, the initial velocity of the police car, u = 100 Km/h = 27.78 m/s
The final velocity of the car, v = 80 Km/h = 22.22 m/s
The displacement of the car, S = 88m
From the third equation of motion: v²- u² = 2aS
(22.22)² - (27.78)² - = 2 × a × 88
a = - 1.58 m/s²
Calculate the time from the first equation of motion: v = u + at
22.22 = 27.78 + (-1.58) × t
t = 3.52 sec
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a service elevator takes a load of garbage, mass 20.7 kg, from a floor of a skyscraper under construction down to ground level, accelerating downward at a rate of 1.7 m/s2. find the magnitude of the force (in n) the garbage exerts on the floor of the service elevator.
The magnitude of the force that the garbage exert on the floor of the service elevator is equal to 238.05 Newtons.
The mass of the garbage in the elevator is 20.7 kg. The elevator is accelerated downward at a rate of 1.7 m/s².
The acceleration due to gravity is g.
As the elevator accelerator downward then the total net acceleration of the elevator in the download direction is
A = (1.7 + 9.8) m/s².
A = 11.5 m/s².
The force that the garbage exert on the floor of the service elevator will be equal to the product of the mass of the garbage and the net acceleration with which the elevator is going downward,
F = 20.7 × 11.5
F = 238.05 N.
So, the force exerted by the garbage on the floor is 238.09 Newton.
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what kind of motion is a uniform circular motion? does the body have an acceleration in a uniform circular motion?
A uniform circular motion is one in which the velocity of the object remains constant. This means that the object is moving with a constant speed and in a circle, without acceleration.
In this type of motion, the acceleration is zero. Acceleration is defined as a vector quantity, which means that it has both magnitude and direction. In this case, both magnitude and direction are zero, so there is no acceleration—the body moves in a straight line at constant speed in a circle.
A body has an acceleration when it is changing its velocity, either increasing or decreasing it. An example would be stopping on a highway, where your foot would push down harder on the gas pedal to increase your speed while going around a curve, then slowing down as you straightened out again.
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you pull a box with a constant force of 250 n. it takes you 20 s to move the box 10 m along a rough surface at a constant speed. what power are you transferring to the system of the box and the rough surface?
Your power usage when pulling the box is 125 Watts. It takes 20 seconds to move a box 10 meter's along a rough surface when applying a continuous 250 n force.
Given: force = 250 N; distance = 10 m; speed = distance/time (10/20); speed = 0.5 m/s
W = work completed, where W = 250*10 = 2500 Joules.
Your labour of 2500 J by pulling the box
Thermal energy produced—considering that the box's surface and interior are both made of wood. Therefore, the kinematic friction coefficient is 0.3.
The thermal energy is therefore 75 N
thermal energy = 75*10 = 750 Joules, or thermal energy = force*d
force = mu
force= 0.3*250.
So pulling the Box produces 750 Joules of thermal power.
Power exerted = force* v
= 250*0.5
=125 Watts
so, The power exerted by you while pulling box is 125 Watts.
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cell phones use radio waves to transmit information. if your cell phone uses a frequency of 1900 mhz, what is the wavelength in meters of the electromagnetic radiation emitted by your phone? (1 hz
The wavelength of the electromagnetic radiation emitted by your phone is 1.578meters.
What is the wavelength of electromagnetic radiation?The name given to the recognized range of electromagnetic radiation is the electromagnetic spectrum (EMS). The frequency decreases from 3×10²⁶ Hz to 3×10³ Hz as wavelengths rise from around 10⁻¹⁸ m to 100 km.
The wavelength is the separation between two wave crests, while troughs have the same wavelength. Frequency, measured in cycles per second (Hz), is the number of vibrations that pass over a certain location in a second (Hertz).
Given,
Frequency = 1900 mhz = 1900×10⁶,
velocity = 3×10⁸,
Wavelength = v/F
Wavelength = 3×10⁸/ 1900×10⁶
Wavelength = 1.578meters.
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What is work and give an example?
Calculate the efficiency of a machine that requires 20 J of input work to do 10 J of output work. What is the efficiency of the machine? (in percent, but write the number only)
The efficiency of a machine that requires 20 J of input work to do 10 J of output work. The efficiency of the machine η = 50 %
The engine efficiency is defined as ratio of the useful work done to the heat provided.
By using the formula for efficiency of a machine and substituting the values in it, we get :
η = (Wo/Wi × 100)%
η = (10/20 × 100)%
η = 50 %
The efficiency of the machine will be η = 50 %
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a satellite is orbiting a planet at distance r above its surface and has a period of t. what would the distance above the surface have to be in order for the period to become eight times greater?
The radius of the new orbit should be 4 times larger to increase the satellite period 8 times.
Calculation:-
Using the formula for the satellite period.
T = 2π [tex]\sqrt{\frac{R^{3} }{GM} }[/tex]
In which R is the average radius of orbit for the satellite G is the gravitational constant, and M is the mass of the planet.
Therefore the radius of the new orbit ought to be four instances large to increase the satellite period 8 instances.
The orbital length is without delay associated with the common distance among the planet and the sun. This regulation means that planetary orbital velocity decreases with increasing distance from the sun.
The orbital length will increase if the orbital distance is multiplied. As the distance of the planet from the solar increases, the length of revolution decreases. you could calculate the space in more than one methods. One way is to multiply the calculated rate by the length.
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a block has accleration a when pulled by a string. if two identical blocks are glued together and pulled with twice the orignal force, their accerleation will be?
you get the same acceleration.
we know that:
F = ma
a = F/m
a = 2F/2m
a = F/m
so if we double the force and the mass, you get the same acceleration.
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a 0.20-kg ball on a stick is whirled on a vertical circle at a constant speed. when the ball is at the three o'clock position, the stick tension is 16 n. find the tensions in the stick when the ball is at the twelve o'clock and at the six o'clock positio
The tension when the stick is at six o clock will be 17.96 N and when the stick will be on 12 o clock the tension will be 14.04 N.
What is tension and how the tensions are calculated out to be so ?Tension is actually the force acting and that both have same unit Newton whose formula is T =mg + ma.Here the question is saying that0.20 kg ball in a stick is whirled on a circle at a constant speed.When the stick would be at 6 position , T6 = mg+T3 = 0.2 x 9.8 +16 = 17.96 N.When the stick would be at 12 position, T12 = T3-mg = 16- o.2x9.8 = 14.04N.Hence when a 0.20 kg ball on a stick whirling on a verticle circle at constant speed , the tension at six and twelve will be 17.96 N and 14.04N respectively.To know more about speed visit:
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a solid ball with a volume of 0.4 m3 is made of a material with a density of 2500 kg/m3. what is the mass of the ball?
1000 kg is the mass of the ball.
How can density be converted from volume?To find the volume of a substance, divide its mass by its density (mass/density = volume). Consistent measurement units should always be used. If the density is specified in grams per cubic centimeter, for instance, then calculate the mass in grams and provide the volume in cubic centimeters.
How can density be determined in three different ways?
From mass and volume measurements, the densities of brass and aluminum will be determined. Three different methods will be used to calculate volumes: geometrically (by measuring lengths); via water displacement; and via pycnometry.
The density calculation formula is p = m/V.
m = p*v
m = 100kg
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What accounts for the light and dark bands produced when monochromatic light reflects from a glass pane atop another glass pane?.
The cause of these bands is the interference between the waves reflected from the glass on the upper and lower surfaces of the air wedge.
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A ball with a mass of 4 kg is dropped from a height of 20 m. What is the velocity of the ball just before it hits the ground? g = 10 m/s^2
Taking into account the definition of potencial energy, the velocity of the ball just before it hits the ground is 20 m/s.
Potential energyPotential energy is the energy that a body has at a certain height above the ground.
Gravitational potential energy is the energy associated with the gravitational force and it depend on the height of an object to some reference point, the mass, and the force of gravity.
The expression applied to the gravitational energy of the object is:
Ep= m×g×h
where:
Ep is the potential energy in joules (J).m is the mass in kilograms (kg).h is the height in meters (m).g is the acceleration of fall in m/s².Kinetic energyKinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is represented by the following expression:
Ec = 1/2× m× v²
where:
Ec is kinetic energy, which is measured in Joules (J).m is mass measured in kilograms (kg).v is velocity measured in meters over seconds (m/s).Principle of conservation of mechanical energyMechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position.
The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative.
Therefore, if the potential energy decreases, the kinetic energy will increase while if the kinetics decreases, the potential energy will increase.
Velocity of the ballIn this case, you know:
Ep= ?m= 4 kgg= 10 m/s²h= 20 mReplacing in the definition of potential energy:
Ep= 4 kg× 10 m/s²× 20 m
Solving:
Ep= 800 J
Considering the principle of conservation of mechanical energy, the ball has 800 J of gravitational potential energy. As it is dropped, this energy will be converted into kinetic energy. Then:
800 J= 1/2× 4 kg× v²
Solving
800 J= 2 kg× v²
800 J÷ 2 kg= v²
400 m²/s²= v²
√400 m²/s²= v
20 m/s= v
Finally, the velocity of the ball will be 20 m/s.
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a heavy brass ball is used to make a pendulum with a period of 5.5 s. part a how long is the cable that connects the pendulum ball to the ceiling? how long is the cable that connects the pendulum ball to the ceiling? 4.7 m 6.2 m 7.5 m 8.7 m request answer provide feedback
The length of the cable that connects the pendulum ball to the ceiling with a period of 5.5 s is 7.5 m
T = 2 π √ ( L / g )
T = Time period
L = Length of the pendulum
g = Acceleration due to gravity
T = 5.5 s
g = 9.8 m / s²
5.5 = 2 * 3.14 √ ( L / 9.8 )
√ L = 0.88 * √ 9.8
L = 0.77 * 9.8
L = 7.5 m
A simple pendulum consists of a mass suspended in air by a massless and unstretchable string that is fixed to a rigid support on the other end. The time taken to complete on oscillation is the time period of a simple pendulum.
Therefore, the length of the cable is 7.5 m
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a stomp rocket is launched straight into the air and is being watched by students 500 meters away. after 2 seconds, when the angle of elevation is the angle is increasing at a rate of 0.4 radians per second. how fast is the stomp rocket rising at that moment?
Using the concepts of pressure, we find that 625m[tex]s^{-1}[/tex] is the speed of the stomp rocket rising at that moment when it is watched by students 500 meters away. after 2 seconds, when the angle of elevation is the angle is increasing at a rate of 0.4 radians per second.
We know that when an object is very far away, at the centre of that object there is an angle enclosed by the observer and that is given
angle=(length of arc)/radius of the arc.
θ=l/r
So, we are given that length of arc is 500 metres and angle is 0.4 radians/sec.
So,=>0.4=500/r
=>r=500/0.4
=>r=1250m
Now we are given the time when it is observed i.e. is 2 seconds.
So, we apply distance speed formula.,
which is Distance=Speed × Time
=>1250=Speed×2
=>Speed=1250/2
=>Speed =625m[tex]s^{-1}[/tex].
Hence, when a stomp rocket is launched straight into the air and is being watched by students 500 meters away. after 2 seconds, and the angle of elevation is the angle is increasing at a rate of 0.4 radians per second, the speed of stomp rocket is 625m[tex]s^{-1}[/tex].
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Suppose a spacecraft is orbiting 3.91* 10^5 meters above the surface of Mars. Calculate the magnitude of Mars' gravitational field and the approximate acceleration due to gravity experienced by the spacecraft (the mass of Mars is 6.39* 10^23 kg and its radius is 3.40* 10^6 m). [gravity, gravitational fields, gravitational acceleration]
The magnitude of Mars' gravitational field is 3.66 N/kg.
The approximate acceleration due to gravity experienced by the spacecraft is 3.66 m/s².
What is the gravitational fields?
The gravitational fields of the Mars is calculated by applying the following equation;
E = F/m = GM/r²
where;
G is universal gravitation constantM is the mass of Marsr is the radius of Mars F is gravitational forceE = (6.626 x 10⁻¹¹ x 6.39 x 10²³) / (3.4 x 10⁶)²
E = 3.66 N/kg
The approximate acceleration due to gravity experienced by the spacecraft is calculated as follows;
F = mg = GmM/r²
g = GM/r²
where;
G is universal gravitation constantM is the mass of Marsr is the radius of Mars F is gravitational forceg is acceleration due to gravityg = (6.626 x 10⁻¹¹ x 6.39 x 10²³) / (3.4 x 10⁶)²
g = 3.66 m/s²
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if the moon is setting at 6 a.m., the phase of the moon must be if the moon is setting at 6 a.m., the phase of the moon must be waning crescent. first quarter. third quarter. new. full.
Considering that sunrise will be visible at 6 A.m, the moon will be in the full moon phase at that time.
Explain about the Full moon?The line formed by the sun, Earth, and moon occurs when the moon has shifted 180 degrees from its new moon position. It is called a full moon when the moon's disc is as close to being completely illuminated by the sun as it can possibly be.
Waning Crescent is the phase of the Moon for today and tonight. The ideal time to watch this phase is immediately before sunrise in the western sky.
A full moon may also cause a reduction in deep sleep and a postponement of the onset of REM sleep, according to some research. Furthermore, several investigations have revealed a minor alteration in cardiovascular conditions during a full moon. Researchers are still looking at how the moon affects many physiological and psychological processes.
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for a charged particle, a constant magnetic field can be used to change for a charged particle, a constant magnetic field can be used to change only the direction of the particle's velocity. only the magnitude of the particle's velocity. both the magnitude and direction of the particle's velocity. none of the above.
For a charged particle, a constant magnetic field can be used to change for a charged particle, only the direction of the particle's velocity .
magnetic force = q(VxB)
which is perpendicular to V and B
so it can't change magnitude only can change direction.
What is magnetic field?A magnetic field is a vector field near a magnet, an electric current, or an alternating electric field where the observed magnetic forces exist. A magnetic field consists of moving electric charges and the internal magnetic moments of elementary particles associated with a fundamental quantum property called spin. The magnetic field and the electric field are both related and are components of the electromagnetic force, which is one of the four forces of nature.
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At what time after the discharge begins is the charge on the capacitor reduced to half its initial value?.
The time it takes for the energy stored in the capacitor to be reduced to half its initial value is 0.00693 seconds.
What is meant by capacitor?A capacitor is a device that stores electrical energy in an electric field by accumulating electric charges on two insulated surfaces. It is a two-terminal passive electronic component. Capacitance is the effect of a capacitor.
The time it takes for the energy stored in the capacitor to be reduced to half its initial value is calculated as follows.
Q=Q₀[tex]e^{-t/RC}[/tex]
the time constant, RC = 10 ms = 0.01 s
Q₀ is the initial charged stored
Q is the final charge of the capacitor
Q₀/2=Q₀[tex]e^{-t/RC}[/tex]
Q₀/2Q₀=[tex]e^{-t/RC}[/tex]
1/2=[tex]e^{-t/RC}[/tex]
㏑(1/2)=-t/RC
㏑(1/2)=-t/0.01
-0.693=-t/0.01
t=o.00693 s
Thus, the time at which the energy stored in the capacitor will reduce to half of its initial value is 0.00693 s.
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The complete question is:
"The capacitor in an RC circuit is discharged with a time constant of 10.0ms. At what time after the discharge begins is the charge on the capacitor reduced to half its initial value."
what minimum height must the sides have for this boat to float in perfectly calm water? mastering physics
The height is 0.32788meters or 31.78 cm.
According to the principle of buoyancy, the weight of a floating body equals the weight of the liquid displaced by its submerged part. Therefore, the ship can float on water. Because its weight is equal to the weight of the water you are trying to displace. Each float displaces the swimming liquid by its own weight. Just like a ship floats freely in the water.
The weight of the container must equal the weight of the volume of water displaced by the container. Displacement is the amount of water a ship displaces. If the buoyancy equals the object's weight, the object will remain buoyant at the current depth. Buoyancy always exists whether an object is floating, submerged, or floating in a liquid. When the boat floats, its weight equals the weight of the volume of water displaced.
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An airplane is flying with a force of 800 N. It experiences a force of air resistance of 40 N and a wind force of 60 N, both acting in the opposite direction that the plane is traveling. What is the net force on the airplane?
The net force on the airplane is 420 N.
What is force?An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.
A spring balance can be used to calculate the Force. Newton is the SI unit of force.
Given parameters:
Force of the airplane, F = 800 N.
Force of air resistance, f₁ = 40 N.
Wind force, f₂ = 60 N.
Net force = ?
According to the question, wind force and force of air resistance are acting opposite direction. As, force of air resistance always acting opposite to the direction of plane flying, wind force is acting along the direction of plane flying.
Hence, net force = F - f₁ + f₂
= ( 800 -40 + 60) N.
= 820 N.
So, net force acting on the plane is 820 N.
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what is the theoretical effect on the period when you change the revolving mass while keeping the radial force and radius the same?
Changing the revolving mass and keeping the radial force same the Period will alter. According to variations in revolving mass.
What is radial force?A force applied in a radial direction toward or away from the centre is referred to as a radial force. When a massed body or fluid element moves in a circular motion, an equilibrium between the centripetal force acting from the inside and the centrifugal force acting from the outside is created in the radial direction.
F = mv²/r
F = Force required to keep the mass in a circle
m = mass
v = velocity of the mass
r = radius of the circle.
Since, v = 2πr/t
Where t = period
π = pie
Substitute the value of v
We have, F = m(2πr/t)²/r
F = 4π²r²m/t²r
F = 4π²rm/t²
If the path radius and radial force are kept constant then mass is directly proportional to period
m ∝ t
Which means if mass increases period will increase as well.
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the angular velocity of a rotating rigid body increases from 500 to 1500 rev/min in 120 s. (a) what is the angular acceleration of the body? (b) through what angle does it turn in this 120 s?
The angular acceleration of the body & through certain angle is 0.87266 rad/sec² and 12566.8rad respectively.
Given -angular velocity of a rotating rigid body increases from 500 to 1500 rev/min .
Time= 120s
To find - Angular acceleration and angle
Principle : We have to convert rev/min to rad/sec .
ωf= ωi + αt
ωf- ωi = αt
α= [ ωf- ωi ]/ t⇒angular acceleration formula
ΔФ=ωi t + αt²/2⇒ angle deviation formula
Calculations-
Conversion of Rev/Min to rad/Sec -
ωi= 500 rev/min
= 500 x 2π rad/rev x 1min/60sec = 52.3598 rad/sec
ωf= 1500 rev/min
=1500 x 2π rad/rev x 1min/60sec = 157.079rad/sec
ωi=52.3598 rad/sec
ωf=157.079rad/sec
Now , to compute angular acceleration- α= [ ωf- ωi ]/ t
= [157.079rad/sec- 52.3598 rad/sec]/120
α=0.87266 rad/sec²
Now to compute angle deviated -ΔФ=ωi t + αt²/2
=52.3598 x 120 + [0.87266 x 120 x120]/2
ΔФ= 12566.8rad
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a 70kg skater from side to side on a curved ramp they start at a height of 5m
at the boot on of the ramp they have the speed of 9.5m/s
on the opposite side of the ramp they only reach a maximum height of 4cm
describe the energy stores and transfers that take place as the skater moves form side to side on the ramp
The velocity of the skater will increase as he moves towards the center of the ramp and consequently the kinetic energy will increase. The kinetic energy will converted into potential energy as he moves towards the maximum height.
What is law of conservation of energy?The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.
As the skater move from side to side on the curved ramp, his kinetic energy will be converted into potential energy.
Mathematically, it be written as;
K.E = P.E
¹/₂mv² = mgh
where;
m is the mass of the skaterv is the speed of the skater at the lowest pointh is the maximum height attainedg is acceleration due to gravityAs the skater moves from the highest point with zero velocity, he will have maximum potential energy which decreases as he moves towards the center of the ramp. The velocity will increase as he moves towards the center of the ramp and consequently the kinetic energy will increase.
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light that is polarized along the vertical direction is incident on a sheet of polarizing material. only 67% of the intensity of the light passes through the sheet. what angle is the axis of the polarizer oriented, with respect to the vertical?
125.06⁰angle is the axis of the polarizer oriented, with respect to the vertical
given,
Only 67.0% of the intensity of the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet.
for the first sheet we have malus law as
I =I₀cos²θ₁
here(given)
I/I₀ = 0.67
so cos²θ₁ =0.67
θ₁ = cos⁻¹√(0.67)
= 35.06⁰
which is the angle the first sheet makes with transmission axis
here given no light passes through the second sheet i.e. both the sheets are in crossed position
so the angle the transmission axis of the second sheet makes with the vertical is
θ₂ = θ₁ +90⁰
=35.06 + 90⁰
=125.06⁰
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a 110 g hockey puck sent sliding over ice is stopped in 15 m by the frictional force on it from the ice. (a) if its initial speed is 6.0 m/s, what is the magnitude of the frictional force? (b) what is the coefficient of friction between the puck and the ice?
The frictional force where k is the unitary force of kinetic friction acting on an object and k is the coefficient of kinetic friction. For the vast majority of surfaces, K is less than S.
Explain about the magnitude of the frictional force?The weight of the object and the roughness of the surface in contact determine how much frictional force is generated. The friction force will grow as the roughness increases.
The weight of the object bearing normal to the surface times the coefficient of friction for the contacting surfaces gives the amount of the friction force. The force's direction is perpendicular to the surface and is in opposition to any forces acting on the box.
The amount of kinetic friction is inversely proportional to the amount of normal force and the degree of roughness between the sliding surfaces. The amount of static friction is inversely proportional to the amount of normal force and the degree of roughness between the sliding surfaces.
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a small car with mass .800kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00m. if the normal force exerted by the track (point b) is 6.00n, what is the normal force on the car when it is at the bottom of the track (point a)?
The normal force on the car when it is at the bottom of the track is 22 N.
A frame revolves in a vertical circle such that its motion at unique points is one-of-a-kind then the motion of the frame is said to be vertical circular movement. recollect an item of mass tied at one give up of a mass m tied at one stop of an int extensible string and whirled in a vertical radius of radius r.
Calculation:-
At the inside of the track
N + mg = mv²/r
6 + 0.8 × 10 = m (v²/r)
14 = 0.8 ( acceleration)
acceleration = 17.5 m/sec² towards centre.
Again,
a = v²/r
ar = v²
v = √ar
v = √17.5 ×5
= 9.35 m/s²
For Normal force
N = mv²/r + mg
N = 0.8 × 1.75 + 0.8 × 10
= 14 + 8
= 22 N
A frame spins in a vertical circle in such a way that its movement at exclusive factors differs from the frame's motion, which is assumed to be vertical circular motion. keep in mind a mass m item wrapped in a stretchable string at one end and whirled in a vertical circle with radius d.
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After energy conversions, you end up with the same total amount of energy as the original amount energy. Which law explains this rule?.
The law that explains after energy conversions, the total amount of energy has the same amount as the original amount of energy is the Law of conservation of energy.
The state law of conservation of energy is that energy cannot be created, the energy cannot be destroyed, and the energy can be converted from one form of energy to another form of energy. Examples like mechanical energy. Mechanical energy is the sum of potential energy and kinetic energy.
Potential energy is the kind of energy stored by an object because of its position.Kinetic energy is the kind of energy that is possessed by an object in motion.If an apple fell from a branch of a tree
The potential energy on the branch is the maximum potential energy.When it fell, the potential energy reduces and changes into kinetic energy.When the apple almost reaches the ground, the potential energy gets smaller and the kinetic energy gets bigger.Before it hit the ground the potential energy is the minimum potential energy, but the kinetic energy reaches the maximum kinetic energy. At every point in its movement, the mechanical energy is always the same amount of energy.Learn more about Law of conservation of energy here: https://brainly.com/question/24772394
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in a high wind situation, the flames of a wildfire will bend away from the direction of the wind and the flame height will be different from the flame length. which is used to describe the fireline intensity?
Flame length is used to describe the fireline intensity.
What is fireline intensity?
The sum of the heat released during a given period of time for each unit length of the fire's edge and the rate at which the fire spreads per unit of ground area is called the heat availability of combustion per unit of ground. Btu/sec/ft of fire front is the fundamental unit of measurement.
The rate of energy or heat emission per unit time per unit length of the fire front is called the fireline intensity (kW/m). From a mathematical perspective, it is equal to the product of the fuel's low heat of combustion (kJ/kg), the amount of fuel used in the flame front (kg/m2), and the linear rate of fire
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