Answer:
12 RPM
Explanation:
I would calculate the distance from the edge to the center, and then multiple that by 3RPM with 2 m/sec
From the edge of a cliff, a 0.55 kg projectile is launched with an initial kinetic energy of 1550 J. The projectile's maximum upward displacement from the launch point is +140 m. What are the (a) horizontal and (b) vertical components of its launch velocity?
Answer:
(a) 38.5m/s
(b) 64.4m/s
Explanation:
First, we can obtain the launch speed from the definition of kinetic energy:
[tex]K=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2K}{m}}\\\\[/tex]
Plugging in the given values, we obtain:
[tex]v=\sqrt{\frac{2(1550J)}{0.55kg}}\\\\v=75.0m/s[/tex]
Now, from the conservation of mechanical energy, considering the instant of launch and the instant of maximum height, we get:
[tex]E_0=E_f\\\\K_0=U_g_f+K_f\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0_x^2\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0^2\cos^2\theta\\\\\implies \cos\theta=\sqrt{1-\frac{2gh_f}{v_0^2}}[/tex]
And with the known values, we compute:
[tex]\cos\theta=\sqrt{1-\frac{2(9.8m/s^2)(140m)}{(75.0m/s)^2}}\\\\\cos\theta=0.513\\\\\theta=59.12\°[/tex]
Finally, to know the components of the launch velocity, we use trigonometry:
[tex]v_0_x=v_0\cos\theta=(75.0m/s)\cos(59.12\°)=38.5m/s\\\\v_0_y=v_0\sin\theta=(75.0m/s)\sin(59.12\°)=64.4m/s[/tex]
It means that the horizontal component of the launch velocity is 38.5m/s (a) and the vertical component is 64.4m/s (b).
A 5.00-kg box is suspended from the end of a light vertical rope. A time dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to v(t)=(2.00m/s2)t+(0.600m/s3)t2. What is the tension in the rope when the velocity of the box is 9.00 m/s?
Answer:
T = 74.3N
Explanation:
We are given;
v(t) = (2.00m/s²)t+(0.600m/s³)t²
So, when v = 9 m/s;
9 = 2t + 0.6t²
0.6t² + 2t - 9 = 0
Solving this quadratic equation,
t = -5.88 or 2.55
We'll pick t = 2.55 s
Now, kinematic acceleration will be the derivative of the acceleration.
Thus, a = dv/dt = 2 + 1.2t
So, acceleration at that time t = 2.55s is; a = 2 + 1.2(2.55) = 5.06 m/s²
Since the rope is subject to both acceleration and gravity, Tension is;
T = mg + ma
T = m(g + a)
T = 5(9.8 + 5.06)
T = 74.3N
In an ac induction motor, the stator produces a rotating magnetic field that induces current in the rotor windings. The rotor current generates a magnetic field in opposition to the magnetic field of the stator, thereby causing the rotor to turn.A. TrueB. False
A. True
According to the Faraday-Lenz law, the current in the rotor is generated by the change in the magnetic flux due to the rotating stator, the polarity of the induced current is such that it produces a magnetic field that opposes the change in magnetic flux that produced it. Which causes the rotor to turn
What is the energy source that heats a contracting protostar?
A current carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire
Explanation:
Given that,
Diameter of the gold wire, d = 0.84 mm
Radius, r = 0.42 mm
Electric field in the wire, E = 0.49 V/m
(a) Electric current density is given by :
[tex]J=\dfrac{I}{A}[/tex]
And electric field is :
[tex]E=\rho J[/tex]
[tex]\rho[/tex] is resistivity of Gold wire
So,
[tex]E=\dfrac{\rhi I}{A}\\\\I=\dfrac{EA}{\rho}\\\\I=\dfrac{0.49\times \pi (0.42\times 10^{-3})^2}{2.44\times 10^{-8}}\\\\I=11.12\ A[/tex]
(b) The potential difference between two points in the wire is given by :
[tex]V=E\times l\\\\V=0.49 \times 6.4\\\\V=3.136\ V[/tex]
(c) Resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=2.44\times 10^{-8}\times \dfrac{6.4}{\pi (0.42\times 10^{-3})^2}\\\\R=0.281\ \Omega[/tex]
Hence, this is the required solution.
Final answer:
To find the current carried by the gold wire, use Ohm's law. To find the potential difference between two points in the wire 6.4 m apart, multiply the electric field by the distance. To find the resistance of a 6.4-m length of the wire, use the formula R = ρL/A.
Explanation:
To find the current carried by the gold wire, we can use Ohm's law which states that the electric field (E) is equal to the product of the current (I) and the resistance (R) of the wire. Rearranging the equation, we can solve for the current: I = E/R. Given that the diameter of the wire is 0.84 mm, we can calculate the cross-sectional area using the formula for the area of a circle (A = πr^2). With the cross-sectional area, we can find the resistance by using the formula R = ρL/A, where ρ is the resistivity of gold and L is the length of the wire. Finally, we can substitute the values into the equation to find the current carried by the wire.
(b) To find the potential difference between two points in the wire 6.4 m apart, we can multiply the electric field by the distance between the two points: V = E × d.
(c) The resistance of a 6.4-m length of this wire can be found using the formula R = ρL/A, where ρ is the resistivity of gold, L is the length of the wire, and A is the cross-sectional area of the wire.
Convection
A) is energy similar to radiant heat from the sun.
B) is the result of conductive heat loss to the air that overlies the surface of the body.
C) is the direct transfer of energy through physical contact.
D) changes a liquid to a vapor.
Answer:
A) energy similar to radiant heat from the sun
A roadway for stunt drivers is designed for racecars moving at a speed of 97 m/s. A curved section of the roadway is a circular arc of 420 m radius. The roadway is banked so that a vehicle can go around the curve with the friction force from the road equal to zero. At what angle is the roadway banked?
Answer:
Explanation:
Given that,
The speed of the car is
Vc = 97m/s
The radius of circular path of the car is
Rc = 420m
We want to find the angle of roadway banked β?
To determine the angle of roadway banked, we will use the formula
tanβ = Vc² / Rc•g
Where Vc = 97m/s, Rc = 420m and
g = 9.8m/s²
Then
tanβ = 97² / (420 × 9.8)
tanβ = 2.28596
β = ArcTan ( 2.28596)
β = 66.37°
The railway banked at an angle of 66.37°
Answer:
Banking angle is 66.35°
Explanation:
Given radius r=420m
Speed=97m/s
banking angle is A
Note before
(V)=√(r*gtanA)
√97=√420*9.81*tanA)taking square of both sides
97^2=420*9.81*tan A.
tanA=66.35°
A=66.35°
A raft with a mass of 180 kg carries two swimmers having masses of 50 kg and 80 kg. The raft is initially at rest. The swimmers dive off opposite ends of the raft at the same time, each with a speed of 3 m/s. At what velocity and direction will the raft move?
Answer:
V = 0.5 m/s
Explanation:
Given,
Mass of the raft,M= 180 Kg
initial velocity of raft,u = 0 m/s
Final velocity of the raft,V = ?
Mass of swimmer 1, m₁ = 50 Kg
initial velocity of swimmer, u₁ = 0 m/s
Final velocity of the swimmer 1, v₁ = 3 m/s
Mass of swimmer 2, m₂ = 80 Kg
Initial velocity of the swimmer 2, u₂ = 0 m/s
Final velocity of the swimmer 2, v₂ = - 3 m/s
Using conservation of momentum
M u + m₁ u₁ + m₂ u₂ = M V + m₁ v₁ + m₂ v₂
M x 0 + m₁ x 0₁ + m₂x 0 = 180 x V + 50 x 3 + 80 x (-3)
180 V = 90
V = 0.5 m/s
Hence, the speed if the raft is 0.5 m/s in the direction of swimmer 1.
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in the bearings causes the wheel to stop in just 13 s . If the moment of inertia of the wheel about its axle is 0.33 kg⋅m2 , what is the magnitude of the frictional torque?
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
What r photons?
........................,........,..........
Answer:
in physics, a photon is a bundle of electromagnetic energy. It is the basic unit that makes up all light
Explanation:
Answer:
The quantum of light and other electromagnetic energy, regarded as a discrete particle having zero rest mass, no electric charge, and an indefinitely long lifetime. It is a gauge boson.
You find it takes 200 J of heat to take 4 kg of an unknown substance from 200 K to 240 K. It does not change phases during this interval. If you added 300 J of heat to the substance instead of 200 J, what would be its final temperature?
Answer:
300 K
Explanation:
First, we have find the specific heat capacity of the unknown substance.
The heat gained by the substance is given by the formula:
H = m*c*(T2 - T1)
Where m = mass of the substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
From the question:
H = 200J
m = 4 kg
T1 = 200K
T2 = 240 K
Therefore:
200 = 4 * c * (240 - 200)
200 = 4 * c * 40
200 = 160 * c
c = 200/160
c = 1.25 J/kgK
The heat capacity of the substance is 1.25 J/kgK.
If 300 J of heat is added, the new heat becomes 500 J.
Hence, we need to find the final temperature, T2, when heat is 500 J.
Using the same formula:
500 = 4 * 1.25 * (T2 - 200)
500 = 5 * (T2 - 200)
100 = T2 - 200
=> T2 = 100 + 200 = 300 K
The new final temperature of the unknown substance is 300K.
) Diameter of a star: Assume that the temperature of the Sun is 5800K and the temperature of Sirius A is 10,000K. If the luminosity of Sirius A is 23 times that of the Sun, calculate the radius of Sirius A relative to the Sun. Sirius A is ______ times the size of the Sun. Your answer should be a number in the format (x.x).
Answer:
Sirius A is 1.608 times the size of the Sun.
Explanation:
The radiant flux establishes how much energy an observer or a detector can get from a luminous source per unit time and per unit surface area.
[tex]R_{p} = \frac{L}{4\pi r^2}[/tex] (1)
Where [tex]R_{p}[/tex] is the radiant power received from the source, L is its intrinsic luminosity and r is the distance.
The Stefan-Boltzmann law is defined as:
[tex]R_{p} = \sigma \cdot T^{4}[/tex] (2)
Where [tex]R_{p}[/tex] is the radiant power, [tex]\sigma[/tex] is the Stefan-Boltzmann constant and T is the temperature.
Then, equation 2 can be replaced in equation 1
[tex]\sigma \cdot T^{4} = \frac{L}{4\pi r^2}[/tex] (3)
Notice that L is the energy emitted per second by the source.
Therefore, r can be isolated from equation 3.
[tex] r^2 = \frac{L}{4\pi \sigma\cdot T^{4}}[/tex]
[tex] r = \sqrt{\frac{L}{4\pi \sigma\cdot T^{4}}}[/tex] (4)
The luminosity of the Sun can be estimated isolating L from equation 3.
[tex]L = (4\pi r^2)(\sigma \cdot T^{4}) [/tex]
but, [tex]r = 696.34x10^{6}m[/tex] and [tex]T = 5800K[/tex]
[tex]L_{Sun} = 4\pi (696.34x10^{6}m)^2(5.67x10^{-8} W/m^{2} K^{4} )(5800K)^{4}) [/tex]
[tex]L = 3.90x10^{26} W[/tex]
To find the luminosity of Sirius A, the following can be used:
[tex]\frac{L_{SiriusA}}{L_{sun}} = 23[/tex]
[tex]{L_{SiriusA}} = (3.90x10^{26} W)(23)[/tex]
[tex]{L_{SiriusA}} = 8.97x10^{27}W[/tex]
Finally, equation 4 can be used to determine the radius of Sirius A.
[tex] r = \sqrt{\frac{8.97x10^{27}W}{4\pi (5.67x10^{-8} W/m^{2} K^{4})(10000K)^{4}}}[/tex]
[tex]r = 1.12x10^{9}m[/tex]
So, Sirius A has a radius of [tex]1.12x10^{9}m[/tex]
[tex]\frac{1.12x10^{9}m}{696.34x10^{6}m} = 1608[/tex]
Hence, Sirius A is 1.608 times the size of the Sun.
Tarzan, whose mass is 94 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.8 m above the ground and the bottom of his dangling feet are at a height 2.0 above the ground. When he first hits the ground he has dropped a distance 2.0, so his center of mass is (2.8 - 2.0) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.5 above the ground.(a) Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground? v = _______ m/s. (b) Consider the extended system. What is the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position?
Tarzan's speed just before his feet touch the ground is 7.4 m/s, calculated using conservation of energy. The net change in internal energy as he bends his knees and stops is 2603.4 J, equivalent to the loss of kinetic energy.
Part (a): Speed of Tarzan Before Touchdown
To find the speed of Tarzan just before his feet touch the ground, we can use the principle of conservation of energy. Initially, Tarzan has gravitational potential energy due to being at a height of 2.8 m above the ground. When he lets go, this potential energy converts to kinetic energy as he falls. At the instant before his feet touch the ground, he is at a height of 0.8 m (2.8 m - 2 m), and the gravitational potential energy at this point converts into kinetic energy. Using the conservation of energy:
Initial Potential Energy = Final Kinetic Energy
mghinitial = 1/2*mv^2
Plugging in the values:
94 kg × 9.8 m/s2 × 2.8 m = (1/2) × 94 kg × v2
After solving the equation, we find:
v = 7.4 m/s
Part (b): Net Change in Internal Energy
The net change in internal energy is the difference between the kinetic energy just before the feet touch the ground and the energy when Tarzan is crouched. Assuming no losses, all the kinetic energy converts into internal energy. Without specific details on work done by Tarzan to flex his muscles or any other form of energy conversion, the kinetic energy that is no longer present in the motion would be assumed to convert fully to internal energy (such as heat). The net change in internal energy equates to the kinetic energy which is lost:
Change in Internal Energy = 1/2*mv^2 - 0 J, since he is at rest.
Using the speed found in part (a) and Tarzan's mass: 1/2*× 94 kg × (7.4 m/s)2
This calculation results in a net change in internal energy of: 2603.4 J
A thin rod of length 0.79 m and mass 130 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 3.61 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
Answer:
a) 0.3965 j
b) 0.3112 m
Explanation:
The picture attached explains it all. Thank you
find the final speed and the time taken for a skier who skies 75 m along a slope that is 15°, measured from horizontal, in the following situations (you may neglect friction). How long (in s) does it take for the skier to reach the bottom of this hill starting from rest?
Answer:
Final speed = 148.21m.s
Time of flight = 7.82seconds
Explanation:
The motion of the body is a projectile motion.
Projectile is a motion created by an object launched in air and allowed to fall to freely under the influence of gravity.
Taking the maximums height reached H = 75m
Angle of launch = 15°
Using the maximum height formula to get the velocity U of the object
H = U²sin²theta/2g
Where g is the acceleration due to gravity = 9.81m/s²
75 = U²(sin15°)²/2(9.81)
1471.5 = U²(sin15°)²
1471.5 = 0.06699U²
U² = 1471.5/0.06699
U² = 21,965.9
U = √21,965.9
U = 148.21m/s
The time taken for the skier to reach the bottom of this hill starting from rest is the time of flight T.
T = 2Usintheta/g
T = 2(148.21)sin15°/9.81
T = 296.42sin15°/9.81
T = 76.72/9.81
T = 7.82seconds
Ultrasound with a frequency of 4.257 MHz can be used to produce images of the human body. If the speed of sound in the body is the same (1.06 km/s) as in salt water, what is the wavelength in the body? Answer in units of m.
The wavelength of an ultrasound with a frequency of 4.257 MHz in the human body is approximately 0.000249 meters or 0.249 mm, calculated using the formula
λ = v/f with the given speed of sound (1060 m/s) and the frequency.
The question is asking to calculate the wavelength of an ultrasound with a frequency of 4.257 MHz in the human body, given the speed of sound in the body is 1.06 km/s. To find the wavelength, we use the formula λ = v/f, where λ is the wavelength, v is the speed of sound, and f is the frequency.
First, we convert the speed of sound from km/s to m/s: 1.06 km/s = 1060 m/s. Then we use the given frequency (f) of 4.257 MHz, which we also convert to Hz: 4.257 MHz = 4.257 x 106 Hz. Now we can calculate the wavelength:
λ = v/f
λ = 1060 m/s / 4.257 x 106 Hz
λ ≈ 0.000249 m
Therefore, the wavelength of the ultrasound in the body is approximately 0.000249 meters or 0.249 mm.
The wavelength of 4.257 MHz ultrasound in the human body, using the speed of sound of 1.06 km/s, is approximately 0.249 mm.
To determine the wavelength of ultrasound in the body given the frequency and the speed of sound, we can use the formula:
lambda = v / f, where
lambda is the wavelength,
v is the speed of sound, and
f is the frequency of the ultrasound wave. The speed of sound in the body is given as 1.06 km/s, which is equivalent to 1060 m/s. The frequency of the ultrasound is 4.257 MHz, or 4.257 x 106 Hz. Plugging these values into the formula:
lambda = 1060 m/s / 4.257 x 106 Hz
The result will yield the wavelength of ultrasound in the human body, measured in meters.
Performing the calculation:
lambda = 1060 / 4.257 x 106
lambda = 0.000249 m or approximately 0.249 mm
The wavelength of 4.257 MHz ultrasound in the human body is therefore about 0.249 mm.
A migrating salmon heads in the direction N 45° E, swimming at 5 mi/h relative to the water. The prevailing ocean currents flow due east at 2 mi/h. Find the true velocity of the fish as a vector. (Assume that the i vector points east, and the j vector points north.)
Answer:
Explanation:
velocity of salmon with respect to water, v(s,w) = 5 mi/h at N 45° E
velocity of water with respect to ground, v(w,g) = 2 mi/h due east
Let the true velocity of salmon is velocity of salmon with respect to water is v(s,g)
First write the velocities in vector from
[tex]\overrightarrow{v}_{s,w}=5(Cos 45\widehat{i}+Sin 45\widehat{j})=3.54\widehat{i}+3.54\widehat{j}[/tex]
[tex]\overrightarrow{v}_{w,g}=2\widehat{i}[/tex]
Using the formula of relative speed,
[tex]\overrightarrow{v}_{s,w}=\overrightarrow{v}_{s,g}-\overrightarrow{v}_{w,g}[/tex]
[tex]3.54\widehat{i}+3.54\widehat{j}=\overrightarrow{v}_{s,g}-2\widehat{i}[/tex]
[tex]\overrightarrow{v}_{s,g}=5.54\widehat{i}+3.54\widehat{j}[/tex]
This i the true velocity of salmon.
The true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
The given parameters;
velocity of the salmon relative to the water, = 5 mi/hdirection of the velocity, θ = 45⁰ocean currents, = 2 mi/hThe true velocity of the fish as a vector is calculated as follows;
[tex]v_f = v\ cos(\theta)i \ + \ v\ sin(\theta)j \ + \ 2i\\\\v_f = 5cos(45) i \ + 5sin(45)j \ + \ 2i\\\\v_f = 3.54i \ + \ 3.54j \ + 2i\\\\v_f = 5.54i \ + \ 3.54 j[/tex]
Thus, the true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].
Learn more here:https://brainly.com/question/4945130
Which best describes an error the students made in the experiment?
Errors in an experiment can include improperly calibrated equipment, human error in measurement timing, and uncontrolled conditions. To reduce these errors, calibration, multiple trials, and consistent variables can be used. Recognizing whether the results are qualitative or quantitative helps in addressing the impacts of these errors.
When considering the question of what best describes an error the students made in the experiment, we can refer to common sources of experimental error. These can vary widely but often include issues with the methods used, errors with the apparatus, or uncontrolled conditions within the experiment. For instance, students may have used improperly calibrated equipment like an electronic balance, leading to systematic errors in measurement. Furthermore, the timing of events using devices like stopwatches can introduce human error if not stopped accurately.
An important thing to consider is whether these errors yield qualitative or quantitative results, as this will determine how we interpret the data. In the case of quantitative results, an inaccurate measurement has a direct numerical impact on the outcome, whereas qualitative errors might inform the observational or descriptive aspects of the results.
To reduce experimental error, students could implement controls such as calibrating equipment before use, running multiple trials to find an average, and ensuring that all variables aside from the independent variable are kept constant. By discussing potential experimental errors and how they might be mitigated, students can better understand how to refine their experimental design for more accurate and reliable results.
Why does sound travel faster in water than in air?
A.
because water is a denser medium than air
B.
because air is a denser medium than water
C.
because water has a larger wavelength than air
D.
because air has a larger wavelength than water
Answer:
C.
Explanation:
Technician A says that to perform a maximum output test on a typical heavy-duty truck charging circuit, the engine should be run at 1500 rpm. Technician B says that alternator rpm is typically at least three times engine rpm. Who is correct?
Answer:
Both Technicaian A and B is correct
Explanation:
An alternator is a device that converts mechanical energy into electrical energy. In every alternator, there is unavoidable power losses. Such losses could be mechanical,drive belt or getting the alternator's bearing heated and electrical loss.more also changing magnetic field also causes some losses in an alternator. The diodes of an alternator get hot when there is drop in votage.At reasonable maximum engine speeds, For instance when it read 1500 RPM to redline and there is heavy loads, starting from the idle through the red line with light electrical accessory loads, Nothing will happen to battery, it will just goes along for the ride. Unless it wait for the alternator to fall below operating speeds.
Therefore, when technician A says that to perform a maximum output test on a typical heavy-duty truck charging circuit, the engine should be run at 1500 rpm, and Technician B says that alternator rpm is typically at least three times engine rpm. Both of them is is correct.
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coasting. Describe each of the following events as an elastic, an inelastic, or a perfectly inelastic collision between the skater and the Frisbee.
Question:
(a) She catches the Frisbee and holds it.
Answer:
The correct option is;
A perfectly inelastic collision
Explanation:
A perfectly inelastic collision is one in which there is maximum amount of loss of kinetic energy in the system. In a perfectly inelastic collision, the colliding members lose their initial speed and they stick together resulting in a loss of kinetic energy.
Since she catches and holds on to the Frisbee, the kinetic energy of the Frisbee is lost as she holds on to it so as to combine her mass to that of the Frisbee.
A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and a cross-sectional area of 9.40 x 10-2 m2 in region B. The pressure difference between the two regions is 6.50 x 103 Pa. What are (a) the volume flow rate and (b) the mass flow rate?
Answer:
A) volume flow rate = 0.047 m3/s
B) mass flow rate = 39.01 kg/s
Explanation:
Detailed explanation and calculation is shown in the image below
what will happen and give a reason
1 - Don't put the coat on the snowman - it will melt him
2 - I don't think the coat will make any difference
3 - I think it will keep him cold and stop him from melting.
Answer:
1 or 2 or 3
Explanation:
it depends whether the coat is a light or dark colour. if it is lighly coloured it will not absorb heat from the sun and wont melt the snowman. if it is darkly coloured it will be number 1
the coat wont make a difference if the temperature of the snowman is the same and the temperature outside the snoman and the sun isnt shining. the coat would just make him look smart - so number 2
it coul keep in the cold and block out the warmth so number 3
If love is the answer, then what is the question?
Answer:
what is happiness
Explanation:
Answer:
Love is the answer to the question of how do you make someone do something that is not to their benefit. How do we stop bad people from doing bad things? How do we overcome greed? How can we find happiness if we always know that we have a limited time on earth?
Explanation:
Un móvil, que sale desde un punto situado 3 metros a la izquierda del origen y lleva un movimiento uniforme, se sitúa a 12 metros a la derecha del origen al cabo de 3 segundos. Tras esto invierte el sentido del movimiento, empleando 4 segundos más en llegar al origen. Realiza una gráfica que represente el movimiento descrito. Obtén el desplazamiento en cada tramo. (Sol: 15 m; −12 m) Determina la distancia total recorrida por el móvil. (Sol: 27 m) Determina la velocidad del móvil en cada tramo. (Sol: 5 m/s; −3 m/s) Representa gráficamente la velocidad del móvil frente al tiempo.
Answer:
x₁ = 15 m, x₂ = 12 m , x_total = 27m, v₁ = 5 m / s , v₂ = - 3 m / s
Explanation:
In this exercise we will use the kinematics of uniform motion
v = d / t
let's apply this equation for the first move
v₁ = Δx / t = (x₂-x₀) / t
v₁ = (12- (-3)) / 3
v₁ = 5 m / s
the distance traveled is x₁ = 15 m
Now let's analyze the return movement
v₂ = Δx / dt
v₂ = (0 - 12) / 4
v₂ = - 3 m / s
The negative sign indicates that the vehicle is moving to the left
the distance traveled is x₂ = 12 m
The total dystonia is
x_total = x₁ + x₂
x_total = 15 +12
x_total = 27m
In the attached we have the graphics of the movement
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating a a steady speed of 10 km/h. Neglecting friction and air drag and assuming the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on. (steady power = 68.1 kW; start-up = 43.7 kW)
Answer:
P = 68.125 kW
P startup = 43.05 kW
Explanation:
The power required to operate this ski lift is 43.05 KW.
What is power?The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units. A scalar quantity is power.
The lift is operating a a steady speed of 10 km/h = 10×(1000/3600) m/s = 25/9 m/s.
Average mass of the chair= 250 kg.
Acceleration of the chair = 25/9 /5 m/s² = 5/9 m/s².
Vertical component of this acceleration is = (5/9) ×{200/√(200²+1000²)} m/s²
= 0.1089 m/s²
Hence, the required height = 1/2 × 0.1089 ×5² m = 1.362 m.
So, the total work done = mgh +1/2mv²
= 250×9.8×1.362 + 1/2× 250 × (25/9)² joule
= 215240.56 joule
So, the power required to operate this ski lift = 215240.56 /5 watt = 43048.112 W = 43.05 KW.
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A mass m = 1.2 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(Ït â Ï). The positive y-axis points upward.a. Find the angular frequency of oscillations in radians per second.b. Determine the value of A in meters.c. Determine the value of Ï in radians.d. Enter an expression for velocity along y axis as function of time in terms of A, Ï and t using the value of Ï from part c.e. What is the velocity of mass at time t = 0.25 s?f. What is the magnitude of mass's maximum acceleration?
Answer:
a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²
Explanation:
Here is the complete question
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y ( t ) = A c o s ( ω t − ϕ ) , with the positive y-axis pointing upward. At time t = 0 the mass is observed to be passing through its equilibrium height with an upward speed of v 0 = 3.9 m/s.
A. Find the smallest positive value of ϕ , in radians.
B. Calculate the value of A in meters.
C. What is the mass's velocity along the y-axis in meters per second, at the time t = 0.15 s?
D. What is the magnitude of the mass's maximum acceleration, in meters per second squared?
Solution
a. Since y ( t ) = A c o s ( ω t − ϕ ), the smallest possible value for ϕ is gotten when c o s ( ω t − ϕ ) = 0 ⇒ ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0, ω t − ϕ = ω 0 − ϕ = 0 − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
Since this is a negative angle, we add 2π to the right side.
So, ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
b. Since v = Aω = A√(k/m) where v = maximum velocity at time t = 0 = 3.9 m/s. A = amplitude, k = spring constant = 130 N/m and m = mass = 1.1 kg
A = v/√(k/m) = 3.9 m/s/√(130 N/m/1.1 kg) = 3.9/√118.18 = 3.9/10.87 = 0.36 m
c. To find its velocity, we differentiate y(t)
So, v = dy(t)/dt = dA c o s ( ω t − ϕ )/dt = -'ωAsin( ω t − ϕ ) = v₀sin( ω t − ϕ )
v = v₀sin( ω t − ϕ ) = v₀sin( ω t − ϕ)
Substituting the value of the variables,
v = 3.9sin( 10.87 t − 3π/2)
At t = 0.15 s,
v = 3.9sin( 10.87 × 0.15 − 3π/2)
v = 3.9sin( 1.6305 − 4.7124)
v = -'3.9sin( -3.0819)
v = -'3.9 × - 0.06
v = 0.234 m/s
d. The maximum acceleration, a
a = Aω² = Ak/m = 0.36 × 130/1.1 = 42.55 m/s²
Answer:
a) F = 10.4 rad/s
b) A = 0.375 m
c) ϕ = 3π/2
d) V(t) = -ωAsin( ω t - 3π/2 )
e) V = 0.144 m/s
f) a = 40.625 m/s²
Explanation: Given that
mass m = 1.2 kg
The spring constant k = 130 N/m Time t = 0
Distance d = 0.35 m
y( t ) = A c o s ( ω t − ϕ )
At time t = 0
Speed of Vo = 3.9 m/s.
a) Find the angular frequency of oscillations in radians per second
W = √(k/m)
2πF = √(k/m)
F = 1/2π√(k/m)
F = 1/2π √(130/1.2)
F = 1.66Hz
1 Henz = 6.28 rad/s therefore,
F = 1.66 × 6.28
F = 10.4 rad/s
b) Calculate the value of A in meters.
V = Aω = A√(k/m)
where V = 3.9 m/s the maximum velocity at time t = 0
A = amplitude
A = v/√(k/m)
A = 3.9/√(130/1.2)
A = 3.9/10.4
A = 0.375 m
c. Determine the value of ϕ in radians
If y( t ) = A c o s ( ω t − ϕ ) We can obtain the smallest possible value of ϕ when c o s ( ω t − ϕ ) = 0
ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0,
ω(0) − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
This is a negative angle, let us add 2π to the right side. So,
ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
d. Enter an expression for velocity along y axis as function of time in terms of A, ϕ and t using the value of ϕ from part c.
To find expression for velocity, we differentiate y(t) with respect to time t So,
V = dy/dt = dA c o s ( ω t − ϕ )/dt
V = -ωAsin( ω t − ϕ )
Therefore
V(t) = -ωAsin( ω t + π/2 )
Or
V(t) = -ωAsin( ω t - 3π/2 ) ...... (1)
e. What is the velocity of mass at time t = 0.25 s?
From equation (1)
V(t) = V₀sin( ω t − ϕ )
Substituting the value of the variables,
V = 3.9sin( 10.4t − 3π/2)
At t = 0.25 s,
V = 3.9sin( 10.4 × 0.25 − 3π/2)
V = 3.9sin( 2.6 − 4.7124)
V= -3.9sin( -2.1124)
V= -3.9 × - 0.037
V = 0.144 m/s
f. What is the magnitude of mass's maximum acceleration?
The maximum acceleration a = Aω²
a = Aω² =
a = Ak/m
a = 0.375 × 130/1.2
a = 40.625 m/s²
what happens when a gas is exposed to a small flame. list as many results.
Answer:
Explanation:
What happens when a flame is deprived of oxygen? Carbon dioxide molecules are heavier than air. Because of this, they push the oxygen and other molecules in the air out of the way as they sink down over the flame and candle. When oxygen is pushed away from the wick, it can't react with the wax anymore. This makes the flame go out.
So if the flame is exposed to air, the flame will rise.
Answer: When gas is exposed to a small flame it explodes into a firerer blow torch at an unbelievable temperature of almost 450 degrees ^F
Explanation: This is just like igniting a lighter then placing an aerosol can of
fumeable gas then spraying the flammeable gas from the aerosol can into the flame of the lighter which will then give you a flamethrower.
_________ is a layer of the earth that is classified not by composition
Answer:
Asthenosphere
Explanation:
The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.
The cytoplasm is the watery fluid found within cells. The cytoplasm holds all of the organelles, except _______, in place within the cell.
A.
chloroplasts
B.
mitochondria
C.
vacuoles
D.
the nucleus
Answer:
The cytoplasm holds all the organelles, except the nucleus.
Explanation:
The cytoplasm holds all organelles, with the primary exception being the nucleus, which is separated by a nuclear envelope. Other organelles like chloroplasts, mitochondria, and vacuoles, are held within the cytoplasm.
Explanation:The cytoplasm, the watery fluid inside cells, houses almost all the organelles. The primary organelle that it does not encompass is the nucleus. The nucleus stays separated from the cytoplasm by a nuclear envelope.
Other organelles such as chloroplasts, mitochondria, and vacuoles are embedded within the cytoplasm. These organelles perform various roles within cells, while the nucleus controls the cell's activities and contains the cell's genetic information.
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