Answer:
d. 4.3 x 10²⁸ kg.
Explanation:
The force that keeps the satellite circling the planet, is the centripetal force, which can be expressed as follows:[tex]F_{c} = m_{s}* \frac{v^{2}}{r} (1)[/tex]
This force is not a different force, is just the net force acting on the satellite towards the planet, which is simply the gravitational force between the planet and satellite masses, that we can write as follows, according the Newton's Universal Law of Gravitation:[tex]F_{g} = G*\frac{m_{s}*m_{p} }{r^{2}} (2)[/tex]
From (1) and (2) as they are equal each other we could get the mass of the planet.However, we don't know the value of the linear speed v, but we know the value of the period T instead.By definition, T is the time needed for the satellite to complete a single circumference of the orbit, so, applying the definition of average velocity, we get:[tex]T = \frac{\Delta s }{v} = \frac{2*\pi *r}{v} (3)[/tex]
Solving (3) for v, replacing the value in (1), and solving finally for mp in (2), we get:[tex]m_{p} = \frac{4*\pi ^{2} *r^{3} }{T^{2}*G} = 4.3e28 kg (4)[/tex]The mass of planet Nutron is ; ( D ) [tex]M_{b} = 4.3 * 10^{28} kg[/tex]
Given data :
Satellite period of circling ( T ) = 84 secs
Radius ( r ) = 8.0 * 10⁶ m
Determine the mass of the planet Nutron
we will apply the equation representing the centripetal force
Centripetal force ( Fc ) = [tex]m_{s} * \frac{v^2}{r} ----- ( 1 )[/tex]
applying Newton's universal law of Gravitation to equation ( 1 )
Fg = [tex]G * \frac{m_{s} * m_{b} }{r^2}[/tex] ------ ( 2 ).
note : equation ( 1 ) equals equation ( 2 )
First step : express T in terms of v and change in position ( Δs )
Given that v ( linear speed ) is unknown
T = Δs / v = [tex]\frac{2\pi r}{v}[/tex] --- ( 3 )
∴ v = [ 2 * π * ( 8 * 10⁶ ) ] / ( 84 )
= 598639.45 m/s
Input result in equation ( 1 ) and solve for the mass of planet Nutron in equation 2
∴ Mass of planet Nutron = [tex]M_{b} = 4.3 * 10^{28} kg[/tex]
Hence we can conclude that The mass of planet Nutron is ; ( D ) [tex]M_{b} = 4.3 * 10^{28} kg[/tex]
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A metal wire is uncharged. Explain how it is possible for a current to flow through it.
Answer:
Most conductors are fully charge neutral when carrying current, since their atomic structure is not altered, and electrons are in their normal state of jumping from atom to atom all the time.
It is just that this movement gets a direction under an external electric field, which causes a net charge flow (current) in conductor. So the conductor remains uncharged.
It is however possible for a capacitor plate to be charged (charged capacitor), and at the same time the plate can be a part of current carrying circuit through two points on it. In this case, charge on plate is independent of current and unrelated with it.
I have made foil type capacitors with such construction where the capacitor was a part of filter circuit in a signal carrier circuit, and connections were made from both ends of each foil.
explain an experiment of the phenomenon of rainfall
Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.
Explanation:
Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.
The water cycle makes rainfall possible:
First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.According to Steinberg, which of the following statements is true about creative people
They appreciate art and music.
They always take popular stands.
They accept ideas at fàce value.
They accept their limitations,
A ball is dropped from rest. Its energy is transformed from ________.
Answer:
gravitational potential energy
When the ball falls, some of its gravitational potential energy is converted to other forms of energy, such as kinetic energy and rotational energy. When it rebounds, these other forms of energy is converted back to gravitational potential energy.
Explanation:
A student throws a baseball upwards at an angle of 60 degrees to the horizontal. The initial horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively. What position will have the greatest magnitude of vertical acceleration
Answer:
All positions have constant acceleration
Explanation:
Projectiles have two velocity component; a vertical velocity and a horizontal velocity. The horizontal velocity is constant while the vertical velocity is acted upon by the force of gravity, causing a change of vertical velocity.
If the air resistance is negligible, the acceleration of a projectile would be constant and equal to the acceleration due to gravity. This means that the acceleration for a projectile is the same at every point in its trajectory.
The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/m. The tension in the string is __________.
Answer:
The tension on the string is 2.353 N.
Explanation:
Given;
the speed of sound in air, v₀ = 343 m/s
then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s
mass per unit length, m/l = μ = 0.002 kg/m
The speed of sound on the string is given as;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu[/tex]
where;
T is the tension on the string
T = (34.3)²(0.002)
T = 2.353 N
Therefore, the tension on the string is 2.353 N.
Help me please,
A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?
A) v 0m/s, = 0m/s/s
B) v = 0m/s, a 10m/s/s
C) v = 10m/s, a 10m/s/s
D) v = 10m/s, a = 0m/s/s
E) None of the above
Option B
Explanation:
no distance was given only the acceleration due to the fact that it went up (10m/s/s)
s0 it is
0 m/s and 10m/s/s (option B)
Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) is at an angle of 5o with respect to the normal.
A. Find the number of slits per centimeter in the grating.
B. Two rays of light of wavelength 650 nm and 420 nm are normally incident on a different grating. If the grating has 5000 slits/cm, what is the angular seperation of of the two light rays' second order maximum?
Answer:
A
[tex]N = 1340.86 \ slits / cm[/tex]
B
[tex]\theta = 15.7^o[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 650 \ nm = 650 *10^{-9} \ m[/tex]
The angle of first bright fringe is [tex]\theta = 5^o[/tex]
The order of the fringe considered is n =1
Generally the condition for constructive interference is
[tex]dsin (\theta ) = n * \lambda[/tex]
=> [tex]d = \frac{1 * 650 *10^{-9 }}{ sin(5)}[/tex]
=> [tex]d = 7.458 *10^{-6} \ m[/tex]
Converting to cm
[tex]d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm[/tex]
Generally the number of grating pre centimeter is mathematically represented as
[tex]N = \frac{1}{d}[/tex]
=> [tex]N = \frac{1}{0.0007458}[/tex]
=> [tex]N = 1340.86 \ slits / cm[/tex]
Considering question B
From the question we are told that
The first wavelength is [tex]\lambda_1 = 650 \ nm = 650 *10^{-9} \ m[/tex]
The second wavelength is [tex]\lambda_2 = 429 \ m = 420 *10^{-9 } \ m[/tex]
The order of the fringe is [tex]n = 2[/tex]
The grating is [tex]N = 5000 \ slits / cm[/tex]
Generally the slit width is mathematically represented as
[tex]d = \frac{1}{N }[/tex]
=> [tex]d = \frac{1}{ 5000 }[/tex]
=> [tex]d = 0.0002 \ c m = 2.0 *10^{-6} \ m[/tex]
Generally the condition for constructive interference for the first ray is mathematically represented as
[tex]d sin(\theta_1) = n * \lambda_1[/tex]
=> [tex]\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}][/tex]
=> [tex]\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}][/tex]
=> [tex]\theta_1 = 40.5 ^o[/tex]
Generally the condition for constructive interference for the second ray is mathematically represented as
[tex]d sin(\theta_2) = n * \lambda_2[/tex]
=> [tex]\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}][/tex]
=> [tex]\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}][/tex]
=> [tex]\theta_2 = 24.8 ^o[/tex]
Generally the angular separation is mathematically represented as
[tex]\theta = \theta_1 - \theta_1[/tex]
=> [tex]\theta = 42.5^o - 24.8^o[/tex]
=> [tex]\theta = 15.7^o[/tex]
A block of mass, m, sits on the ground. A student pulls up on
the block with a tension, T, but the block remains in contact
with the ground. What is the normal force on the block?
Answer a
Explanation: a
How would the mass and weight of an object on the Moon compare to the mass and weight of the same object on Earth? * Mass and weight would both be less on the Moon. Mass would be the same but its weight would be less on the Moon. Mass would be less on the Moon and its weight would be the same. Mass and weight would both be the same on the Moon.
Answer:
B. Mass would be the same but its weight would be less on the Moon.
Explanation:
The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.
The appropriate option is B.
The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.
W = mg
Here, W is weight, m is mass and g is gravitational acceleration.
Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.
Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
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PLEASE HELP :(
I WILL GIVE EXTRA POINTS
1. When is the kinetic energy of an electron transformed into potential energy?
when it interacts with other electrons, decreasing its speed
when it interacts with neutrons without changing its speed
when it interacts with neutrons , increasing its speed
when it interacts with other electrons without changing its speed
2. Atoms bond to form molecules. Which structures or regions of the atoms interact in bonds ?
electric fields of particles with positive charge
electric fields of particles with no charge
electric fields of particles with negative charge
electric fields of particles with opposite charges
3. If two electrons that are apart get pushed toward each other, how does the repulsion between them change?
Initial repulsion is low and decreases as they approach .
Initial repulsion is high and decreases as they approach .
Initial repulsion is high and increases as they approach .
Initial repulsion is low and increases as they approach .
4. A positive charge of 5.0x10 ^ -5 C °is 0.040 m from a second positive charge of 2.0x10 ^ -6 C Calculate the force between the charges.
5.6x10^2 N
5.6x10^2 N
1.4X10^-2 N
2.3X10^1 N
(a) When the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed.
(b) The region of atoms that interact in bonds is electric fields of particles with negative charge.
(c) Initial repulsion is low and increases as they approach.
(d) The force between the charges is 562.5 N.
Kinetic theory of matter
This theory states that, the collision of particles (electrons) of matter is perfectly elastic. This implies that as the particles (electrons) collides with one another, kinetic energy is transferred from one electron to another.
ΔK.E = ΔP.E
Change in kinetic energy is equal to change in potential energy of the electrons.
Thus, when the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed. Decrease in speed implies decrease in kinetic energy and increase in potential energy.
Chemical bonds of moleculesChemical bond is formed from the transfer or sharing of electrons between atoms. (electrons between atoms implies negative charge to negative charge)
Thus, the region of atoms that interact in bonds is electric fields of particles with negative charge.
Coulomb's law
This law states that the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
When the distance between the electrons are large, the repulsive force is low and the distance is small, the repulsive force is high.
Force between the chargesThe force between the charges is determined by applying Coulomb's law,
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{9\times 10^9\times (5 \times 10^{-5}) \times (2\times 10^{-6})}{0.04^2} \\\\F = 562.5 \ N[/tex]
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What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump?
COP =
Complete Question
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What is the refrigerator's coefficient of performance? COP
(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP
Answer:
a
[tex]COP = 8.49[/tex]
b
[tex]COP_1 = 9.49[/tex]
Explanation:
From the question we are told that
The lower operation temperature of refrigerator is [tex]T_1 = -8.00^oC = 265 \ K[/tex]
The upper operation temperature of the refrigerator is [tex]T_2 = 23.2 ^oC = 296.2 \ K[/tex]
Generally the refrigerators coefficient of performance is mathematically represented as
[tex]COP = \frac{T_1}{T_2 - T_1 }[/tex]
=> [tex]COP = \frac{265}{296.2 - 265 }[/tex]
=> [tex]COP = 8.49[/tex]
Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as
[tex]COP_1 = \frac{T_2}{ T_2 - T_1}[/tex]
=> [tex]COP_1 = \frac{296.2}{ 296.2 - 265 }[/tex]
=> [tex]COP_1 = 9.49[/tex]
The coefficient of performance if the refrigerator is used as a heat pump is 9.5.
The given parameters;
initial temperature, T₁ = -8 ⁰C = -8 + 273 = 265 Kfinal temperature, T₂ = 23.2°C = 23.2°C + 273 = 296.2 KThe coefficient of performance if the refrigerator is used as a heat pump is calculated as follows;
[tex]COP = \frac{T_2}{T_2 - T_1}\\\\COP = \frac{296.2}{296.2 - 265} \\\\COP =9.5[/tex]
Thus, the coefficient of performance if the refrigerator is used as a heat pump is 9.5.
"Your question is not complete, it seems to be missing the following information";
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump?
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What is the steady state rate of heat flow through a pane of glass that is 40.0 cm by 30.0 cm with a thickness of 4.00 mm when the outside temperature of the glass is −10.0℃ and its inside temperature is 25.0℃? The thermal conductivity of glass is 0.105 W/(m⋅K); the specific heat of glass is 0.180 cal/(g⋅℃); and 1 cal = 4.190 J.
Answer:
110.25 watt/s
Explanation:
The rate of heat transfer is given by the formula
R = k*A(T1 - T2) / d, where
k = thermal conductivity, 0.105 (W/(m*K)
d = thickness, 4 mm = 0.004 m
T1 = inside temperature, 25° C
T2 = outside temperatures, -10° C
A = area 0.3 * 0.4 = 0.12 m²
Now, applying the given data to the formula, we have
Rate = 0.105 * 0.12 (25 - -10) / 0.004
Rate = 0.105 * 0.12(35) / 0.004
Rate = 0.105 * 4.2 / 0.004
Rate = 0.441 / 0.004
Rate = 110.25 Watt/s
Therefore, the steady state rate of heat flow is 110.25 watt/s
A bicycle has a momentum of 36 kg* m/s and a very!I city of 4 m/s.What is the mass of the bicycle?
p = 36 kgm/s
v = 4m/s
we know that,
p = mv
so,
[tex]m = \frac{p}{v} [/tex]
[tex]m = \frac{36}{4} [/tex]
[tex]m = 9kg[/tex]
If your heart is beating at 76.0 beats per minute, what is the frequency of your heart's oscillations in hertz?A) 3.98 HzB) 4560 HzC) 2.54 HzD) 1450 HzE) 1.27 Hz
Answer:
E) 1.27 Hz
Explanation:
If the heart beats 76 times in a minute, this means that the time needed for a single beat (the period of the oscillation), is just 1/76 of a minute, as follows:[tex]T = \frac{1}{76} min = 0.013 min = 0.79 sec. (1)[/tex]
The frequency of the oscillation, by definition, is just the inverse of the period T, as follows:[tex]f = \frac{1}{T} = \frac{1}{.79sec} = 1.27 Hz (2)[/tex]
The right answer is the E).a hiker walks westward 2 meters and then eastward 7 meters. for this motion the distance moved is?
Answer:
d = 9 [m]
Explanation:
This is a problem where we must be clear about the concept of distance, which tells us that it is a measure of the space traveled on a trajectory.
Therefore:
d = 2 + 7
d = 9 [m]
The directions are not taken into account, as it is the sum of the displacements traveled, regardless of their directions.
The size, shape, and color of an object would be an example of a
O physical change
O physical property
O chemical change
chemical property
Answer:
physical change
Explanation:
its only changing things you see.
Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
A string of length 10.0 m is tied between two posts and plucked. This sends a wave down the string moving at a speed of 130 m/s with a frequency of 215 Hz. How many complete wavelengths of this wave will fit on the string?
Answer:
16.
Explanation:
In any wave, by definition, there exists a fixed relationship between the speed v, the frequency f , and the wavelength λ, as follows:[tex]v = \lambda * f (1)[/tex]
In our case, v = 130 m/s and f= 215 Hz, so solving for λ in (1), we get:[tex]\lambda = \frac{v}{f} = \frac{130m/s}{215 hz} = 0.61 m (2)[/tex]
In order to know how many wavelengths of this wave will fit on the string, we need just do divide the length of the string (10.0 m) over one single wavelength, as follows:[tex]n = \frac{L}{\lambda} = \frac{10.0m}{0.61m} = 16.4 (3)[/tex]
Since we need to take the integer value from this expression, the number of complete wavelengths that will fit on this string is just 16. Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N
B. 1.67 X 10-11 N
C. 8.17 X 10-6N
D. 5.78 X 10-6 N
Answer: B
Explanation: the teacher just told us the answer
The gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
The given parameters;
mass of each sphere, m = 1.00 kgdistance between their center mass, r = 2 mThe gravitational force between the two spheres is determined by applying Newton's law of universal gravitation as shown below;
[tex]F = \frac{Gm_1 m_2 }{r^2} \\\\[/tex]
where;
G is universal gravitation constant = 6.67 x 10⁻¹¹ N/m[tex]F = \frac{(6.67\times 10^{-11})\times (1\times 1)}{2^2} \\\\F = 1.67 \times 10^{-11} \ N[/tex]
Thus, the gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
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Two motorcycles are traveling due east with different velocities. However, 3.63 seconds later, they have the same velocity. During this 3.63-second interval, motorcycle A has an average acceleration of 4.55 m/s2 due east, while motorcycle B has an average acceleration of 18.9 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 3.63-second interval, and (b) which motorcycle was moving faster
Answer:
52.095 m/s
Motorcycle a was moving faster
Explanation:
We start by using one of the equations of motion
V = u + at
If the first motorcycle starts with an initial speed of u(a) and accelerates at a value of a(a) = 4.55 m/s², then the final speed after a time of 3.63 seconds is V(a). We then represent it as
V(a) = u(a) + a(a).t
If the second motorcycle starts with an initial speed of u(b) and accelerates at a value of a(b) = 18.9 m/s², then the final speed after a time of 3.63 seconds is V(b). We then represent it as
V(b) = u(b) + a(b).t
Assuming that the final speeds v(a) = v(b), and then subtract the equation of the second motorcycle from that of the first, we have
0 = u(a) - u(b) + a(a).t - a(b).t
-u(a) + u(b) = a(a).t - a(b).t, on rearranging, we have
u(b) - u(a) = [a(a) - a(b)]t
Since we have the values for acceleration and the time, we substitute so that
u(b) - u(a) = (4.55 - 18.9)3.63
u(b) - u(a) = -14.35 * 3.63
u(b) - u(a) = -52.095, or we rearrange to get
u(a) - u(b) = 52.095 m/s
A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.
Given that,
Mass of the object, m = 1.2 kg
Initial speed of the object, u = 8 m/s
Final speed of the object, v = -6 m/s (in opposite direction)
Time, t = 2 ms
To find,
The average force on the object by the wall.
Solution,
Let F be the force. Using Newton's second law of motion,
F = ma, a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]
or
F = 8.4 N
So, the magnitude of average force in the object by the wall is 8.4 N.
Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a force of 325 N to the right. What is the net force on the table?
Answer: 75 N to the right
Explanation:
When two forces acting in opposite directions, they will cancel in magnitudes. Here, the net force acting on the table is 75 N to the right.
What is force?Force is an external agent acting on an object to change its motion or to deform it. Force is a vector quantity. Hence, it is characterized by a magnitude and direction.
If two equal forces are acting on object from the same direction, they will add up in magnitudes and the net force is their sum. If the two forces are from opposite directions, they will cancel each other in magnitude and the net force will be the substracted value.
Here, Bill is pulling by 250 N to the left and Janet is pulling to the right by 325 N. The force is not balanced because they are unequal.
net force = 325 - 250 = 75 N
The larger force is to the right. Hence, the net force is 75 N to the right.
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A woman pulls a 7.87 kg suitcase,
initially at rest, with a 29.2 N force
along the handle at 62.7° Friction
pulls back at 11.7 N, and the
suitcase moves 8.44 m. What is
the total work done?
(Unit = J)
Answer:
Effective Pulling Force = 29.2(cos 62.7°) = 13.393 N
Resistance to Pulling Force = 11.7 N
Net Force acting on suitcase = 13.393 - 11.7 = 1.693 N
Distance moved = 8.44 m
Net Work = 1.693(8.44) = 14.289 ≈ 14.3 J ANS
Explanation:
hope that makes sense homie
Answer:
14.3
Explanation:
right answer for acellus
A carousel at the local carnival rotates once every 45 seconds.
(a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?
Answer:
We know that the carousel does a complete rotation in 45 seconds.
Then the frequency of this carousel will be f = 1/45 seconds.
And the angular frequency will be 2*pi times the frequency, then we have:
angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1
Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:
S = R*W
then:
a) in this case the radius is 2.75m, then the linear speed is:
S = 2.75m*0.1396 s^-1 = 0.3839 m/s
b) in this case the radius is 1.75m, then the linear speed here is:
S = 1.75m*0.1396 s^-1 = 0.2443 m/s
(a) The linear speed of an outer horse on the carousel is 0.384 m/s.
(b) The linear speed of an inner horse on the carousel is 0.244 m/s.
Given data:
The time interval for the rotation of carousel is, t = 45 s.
The distance of the outer horse from the axis of rotation is, r = 2.75 m.
The distance of an inner horse from the axis of rotation is, r' = 1.75 m.
(a)
The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,
v = 2 π r/t
Solving as,
v = 2 π (2.75) / 45
v = 0.384 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.
(b)
Now similarly the linear speed of an inner horse is calculated as,
v' = 2 π r' / t
Solving as,
v' = 2 π (1.75) / 45
v' = 0.244 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.
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A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 7020 W. How much energy does it deliver into a home during 1 h of continuous operation?
Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ
If an atom contains 13 protons, then it has (2.4)a.13 electrons. b. 26 electrons. c. 13 neutrons. d.26 neutrons.
If an atom contains 13 protons, then it has 13 electrons.
A 2150 kg car, moving east at 10.0 m/s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m/s. What is the 3250 kg car’s initial velocity calculated to the nearest tenth? Record your answer in the boxes below. Be sure to use the correct place value.
Answer:
2.1 m/s
Explanation:
According to law of conservation of momentum;
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the common velocity
Given
m1 = 2150kg
m2 = 3250kg
u1 = 10.0m/s
u2 = ?
v = 5.22m/s
Substitute and get u2
2150(10) + 3250u2 = (2150+3250)5.22
21,500 + 3250u2 = 5400(5.22)
3250u2 = 28,188 - 21500
3250u2 = 6688
u2 = 6688/3250
u2 = 2.1 m/s
Hence the 3250 kg car’s initial velocity has an initial velocity of 2.1 m/s
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If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m
PLEASE HELP ME ASAP!! GUYSSS!! I AM IN CLASS AND DYING! LITERALLY
Billy and Ashley live in the same time zone. Billy lives in Brazil (blue smiley face on the image below). Ashley lives in Eastern Canada (yellow smiley face on the image below).
One day, Billy and Ashley are both outside at 1:32 pm. They are talking on the phone to each other. As they talk, Billy notices the sky in Brazil getting progressively darker. Eventually, it feels like it is nighttime, because it is so dark. Billy thinks the world is coming to an end. He asks Ashley if she is experiencing the same thing in Canada. Ashley has no idea what he is talking about. “It’s perfectly bright and sunny where I am,” she says.
Ashley and Billy conclude that the world is not coming to an end. They reach out to some 7th graders to figure out what is happening. The 7th graders tell Billy that he is experiencing a solar eclipse. To help Billy and Ashley understand, create a model to show why Billy is experiencing an eclipse in Brazil, but Ashley is not experiencing the eclipse in Canada.
Your model must include:
The sun, the earth, the moon, and solar energy (clearly labeled).
How accurate your scale is.
How solar energy interacts with both the moon and with Earth.
The tilt of the moon’s orbit relative to the Earth’s orbit
Why Billy is experiencing the solar eclipse and why Ashley is not.
Answer:
sounds like a you problem
Explanation:
yeah