The correct answer for the (A) Escape velocity is [tex]570[/tex] (B) Escape velocity is [tex]0.57[/tex] in Km/h and (c). Escape velocity is [tex]1.27[/tex] in mph.
Given:
Diameter of asteroid D = [tex]10[/tex] km
Radius R = [tex]5[/tex] Km
Density [tex]\rho[/tex] = [tex]3000[/tex] kg/m³
Unit conversion;
[tex]1[/tex] m/s = [tex]0.001[/tex] Km/s
[tex]1[/tex] m/s = [tex]2.23694[/tex] mph
(A)To calculate Escape velocity:
Use the formula;
[tex]v_e = \sqrt{\dfrac{2GM}{R} }[/tex]
Gravitational Constant [tex]G[/tex] = [tex]6.67430[/tex]
To calculate Mass([tex]M[/tex]) of the asteroid, Calculate Volume([tex]V[/tex]) of the sphere and multiply it with density([tex]\rho[/tex]).
[tex]V= \dfrac{4}{3} \pi R^3 \\\\\rho = \dfrac{M}{V}[/tex]
[tex]M = \rho*V[/tex]
= [tex]523598775000[/tex] Kg
Escape velocity:
[tex]v_e = \sqrt{\dfrac{2*6.67430 * 10^{-11} * 523598775000}{5000} }[/tex]
[tex]= 570[/tex] m/s
(B)Escape velocity in Km/s:
[tex]v_e = \dfrac{570}{1000}[/tex]
[tex]= 0.57[/tex] Km/s
(B)Escape velocity in mph:
[tex]v_e = 0.57 * 2.23694[/tex]
[tex]= 1.27[/tex] mph
Escape velocity is [tex]570[/tex] m/s. In Km/h is [tex]0.57[/tex] and In mph is [tex]1.27[/tex] .
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find the current in an 8.00-v resistor connected to a battery that has an internal resistance of 0.15 v if the voltage across the battery (the terminal voltage) is 9.00 v. (b) what is the emf of the battery?
(a) The flowing current is 1.08 A. (b) The EMF of the battery is 9.16 V.
It is given data that the resistance of the resistor (R) = 8.00 V and the voltage across the battery (V) = 9.00 V. The internal resistance of the battery (r) = 0.15 V
Formula used:
V = EMF - I * rV = IR
Where, V is the terminal voltage of the battery, EMF is the electromotive force of the battery, I is the current flowing through the circuit, and R is the resistance of the resistor. r is the internal resistance of the battery
(a) The current flowing through the circuit can be calculated using the Ohm's Law.
V = IR
I = V / R
I = 9 / (8 + 0.15)
I = 1.08 A
The current flowing through the circuit is 1.08 A.
(b) Find the emf of the battery:
We know that,
V = EMF - I * r
EMF = V + I * r
EMF = 9 + 1.08 * 0.15
EMF = 9.16 V
The emf of the battery is 9.16 V.
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The types of energy in a wave come from the ______ of the wave (potential) and the ______ of the water particles in their orbits (kinetic).
The types of energy in a wave come from the elevation of the wave (potential) and the motion of the water particles in their orbits (kinetic).
What is energy?Energy is the ability to do work. The energy of a wave is measured by its amplitude or wave height. The more energy a wave has, the higher its amplitude. The energy of the wave is the sum of the potential energy and kinetic energy of the water molecules that make up the wave.
What is the potential energy of a wave?A wave has potential energy, which is the energy it possesses due to its position. When a wave is high, it has a lot of potential energy, which can be used to do work. Potential energy is converted to kinetic energy when the wave moves.
What is kinetic energy in a wave?The water particles that make up the wave are in motion. This motion is referred to as kinetic energy. The energy is generated when the wave is in motion. The faster the wave moves, the more kinetic energy it has. Kinetic energy is converted to potential energy when the wave is at its peak.
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An unmanned satellite orbits the earth with a perigee radius of 10,000 km and an apogee radius of 100,000 km. Calculate (a) the eccentricity of the orbit; (b) the semimajor axis of the orbit (kilometers); (c) the period of the orbit (hours); (d) the specific energy of the orbit (kilometers squared per seconds squared); (e) the true anomaly at which the altitude is 10,000 km (degrees); (f) vr and v⊥ at the points found in part (e) (kilometers per second); and (g) the speed at perigee and apogee (kilometers per second).
The speed (at perigee) is 6.32 km/s and the speed at apogee is, 2.00 km/s.
How to calculate the speeds?a) To calculate the eccentricity of the orbit, you need to calculate the ratio of the difference between the perigee and apogee radii to the sum of the two radii. This can be calculated using the following equation:
eccentricity = (apogee radius - perigee radius) / (apogee radius + perigee radius).
Plugging in the given values, the eccentricity of the orbit is,
e = (100,000 - 10,000) / (100,000 + 10,000) = 0.91.
b) The semimajor axis of the orbit (in kilometers) can be calculated using the following equation:
semimajor axis = (perigee radius + apogee radius) / 2.
Plugging in the given values, the semimajor axis of the orbit is,
a = (10,000 + 100,000) / 2 = 55,000 km.
c) The period of the orbit (in hours) can be calculated using the following equation: period = 2π√(semimajor axis^3 / μ). Here, μ is the standard gravitational parameter which is 398,600 km^3/s^2. Plugging in the given values, the period of the orbit is T = 2π√(55,000^3 / 398,600) ≈ 25,000 hrs.
d) The specific energy of the orbit (in kilometers squared per second squared) can be calculated using the following equation: specific energy = (2 / perigee radius) - (1 / semimajor axis).
Plugging in the given values, the specific energy of the orbit is,
specific energy = (2 / 10,000) - (1 / 55,000) ≈ -0.0054 km^2/s^2.
e) The true anomaly at which the altitude is 10,000 km (in degrees) can be calculated using the following equation: true anomaly = arccos((perigee radius / semimajor axis) - 1). Plugging in the given values, the true anomaly at which the altitude is 10,000 km is v = arccos((10,000 / 55,000) - 1) ≈ 74.4°.
f) The radial velocity (vr) and the transverse velocity (v⊥) at the points found in part (e) (in kilometers per second) can be calculated using the following equations:
vr = √(μ / perigee radius) * cos(true anomaly) and v⊥ = √(μ / perigee radius) * sin(true anomaly). Plugging in the given values, the radial velocity at the point of 10,000 km altitude is vr = √(398,600 / 10,000) * cos(74.4°) ≈ 4.17 km/s and the transverse velocity is v⊥ = √(398,600 / 10,000) * sin(74.4°) ≈ 4.04 km/s.
g) The speed at perigee and apogee (in kilometers per second) can be calculated using the following equation:
speed = √(μ/ radius).
Plugging in the given values, the speed at perigee is,
speed (at perigee) = √(398,600 / 10,000) ≈ 6.32 km/s
The speed at apogee is,
speed (at apogee) = √(398,600 / 100,000) ≈ 2.00 km/s.
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what diameter should the nichrome wire be in order for the electric field strength to be the same in both wires?
The diameter of the nichrome wire should be 0.71 times the diameter of the copper wire in order for the electric field strength to be the same in both wires.
To determine the diameter ratio of nichrome wire to copper wire, we can use the equation for electric field strength (E = V / d), where V is the potential difference and d is the distance between the wires.
Since the potential difference between the wires is the same for both wires, we can set the electric field strength for the two wires equal to each other: E_nichrome = E_copper
Using the equation for electric field strength, we can rewrite this as: V_nichrome / d_nichrome = V_copper / d_copper
We know that the diameters of the wires are inversely proportional to their respective distances from each other, so we can rewrite this equation as: V_nichrome / (k * d_copper) = V_copper / d_copper, where k is the ratio of the diameter of the nichrome wire to the diameter of the copper wire.
Solving for k, we get: k = (V_nichrome / V_copper)^(1/2)
Substituting the given values, we get: k = (60 / 40)^(1/2) = 0.866
Therefore, the diameter of the nichrome wire should be 0.71 times the diameter of the copper wire (since 1 / 0.866 = 1.155, and 1.155 * diameter of copper wire = diameter of nichrome wire).
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A weight is connected to a spring that is suspended vertically from the ceiling. If the weight is displaced downward from its equilibrium position and released, it will oscillate up and down.(a) If air resistance is neglected, will the total mechanical energy of the system (weight plus Earth plus spring) be conserved?YesNo(b) How many forms of potential energy are there for this situation?both gravitational and elastic potential energyonly elastic potential energy There is no potential energy in this situation.only gravitational potential energy
a) The mechanical energy of a system is conserved if air resistance is ignored. (b) For this situation, two types of potential energy exist: gravitational potential energy and elastic potential energy.
Explanation: If air resistance is not taken into consideration, the system will be in a state of total mechanical energy conservation. In the absence of air resistance, the kinetic energy and potential energy of the system remain constant, and the mechanical energy remains unchanged.
b) Both gravitational and elastic potential energies are two types of potential energy for this situation. Potential energy is the amount of energy stored in an object as a result of its location or configuration. It may also be stored in a system of objects, like a weight linked to a spring that is suspended from the ceiling vertically.
In a vertical direction, the weight has gravitational potential energy due to its position in the gravitational field of the Earth. The weight is at a specific height from the ground, and this height contributes to the object's potential energy.
The potential energy of a weight suspended from a spring is the second type of potential energy in this scenario. When the spring is stretched, it stores energy in the form of elastic potential energy. The spring's potential energy is transformed into kinetic energy as it vibrates up and down.
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During a baseball game, the sound of the bat hitting the ball can be heard in most parts of the stadium. That sound is weaker at greater distances. What is the cause of this phenomenon?(1 point)
The sound waves are spread out over a large area.
The sound waves are blocked by people in the stadium.
The sound waves can only travel through certain materials.
The sound waves slow down as they move away from the bat.
The cause of the phenomenon is because the sound waves are spread out over a large area. Option 1 is correct.
How do sound waves propagate in this case?In this case, the sound waves are spread out over a large area. This is due to the fact that sound waves propagate outward from their source in all directions, creating a spherical wavefront that expands as it moves away from the source. As the wavefront expands, the same amount of sound energy is distributed over a larger and larger area, causing the sound intensity to decrease with distance from the source. This is known as the inverse square law, which states that the intensity of a sound wave decreases proportionally to the square of the distance from the source. Therefore, the farther away you are from the source of the sound, the weaker it will be.
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Categorize the following exercises as being isometric or isotonic.
Pushing constantly against a concrete wall
Running up a hill
Swimming freestyle
Pedaling a bicycle on a flat surface
Holding a bench-press bar in the same position
Doing a plank exercise (holding a push-up position)
Balancing on tiptoes
Doing bicep curls
Isometric pushes against a wall made of concrete, Isotonic running up a hill. isotonic freestyle swimming, bicycle pedalling on a level surface: isotonic.
Static muscle contractions, in which the length of the muscle does not change during the workout, are called isometric exercises. This indicates that during the activity, there is no discernible movement or alteration in joint angle. Instead, the muscles are tense against a constant force or maintained still for a certain period of time. Exercises that are isometric include pushing against a wall, keeping a plank position, and tightening a hand grasp. Exercises that are isometric can help to increase joint stability and balance as well as muscular strength and endurance. They can also be incorporated into normal workout routines for general health and strength training. They are frequently used in physical therapy to aid patients in recovering from injuries or surgery.
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the bottom of the tank is 30 feet in diameter and groundwater level is 10 ft above the bottom of the tank. how much force is being exerceted on the bottom of the tank?
The bottom of the tank is 30 feet in diameter and the groundwater level is 10 ft above the bottom of the tank. This means that the total force being exerted on the bottom of the tank is equal to the pressure of the water at 10 ft depth, multiplied by the area of the tank.
The pressure of water at 10 ft depth can be calculated using the formula: Pressure = Density of Water x Gravity x Depth
Therefore, the force exerted on the bottom of the tank is:
Force = Density of Water x Gravity x Depth x Area of the TankWhere:
Density of Water is 62.4 lb/ft³Gravity is 32.2 ft/s2, Depth is 10 ftArea of the Tank is 706.5 ft²Therefore, the total force exerted on the bottom of the tank is: Force = 62.4 x 32.2 x 10 x 706.5 = 1358561.2 lbf.
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f the Sun were the size of a small exercise ball (about one-half meter in diameter) and if Jupiter were the size of a golf ball, how big would Earth be on this scale? The size of a hot-air balloon, because Earth is larger than the Sun. The size of a golf ball, because Earth is about the same size as Jupiter. The size of a baseball, because Earth is larger than Jupiter. The size of a pea, because Earth is smaller than Jupiter.
Earth would be the size of a pea on a scale where the Sun is the size of a little exercise ball and Jupiter is the size of a golf ball since Jupiter is significantly larger than Earth.
In this scale, Earth would be the size of a pea if the Sun were the size of a small exercise ball and Jupiter was the size of a golf ball. This is because Earth, which has a diameter of about 12,742 kilometres compared to the Sun's diameter of around 1.4 million kilometres and Jupiter's diameter of approximately 140,000 kilometres, is far smaller than both the Sun and Jupiter. Because of their enormous proportions, celestial bodies' relative sizes in the cosmos might be difficult to comprehend, but making comparisons like these can help put things into perspective and further comprehension.
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When two metal spheres are connected by a metal wire?
The charge is shared equally between the two spheres because metals are good conductors of electricity.
When two metal spheres are connected by a metal wire, the charge is distributed equally between the two spheres. This occurs because metals are good conductors of electricity, which allows electrons to flow freely between them.
The electrons will move from one sphere to the other, redistributing the charge until their charges are equal. This is because of the principle of electric charge distribution, which states that a conductor will always redistribute electric charge until it reaches equilibrium.
The process of connecting two metal spheres with a wire and allowing the electrons to flow between them is an example of electrical conduction.
This is a fundamental process in electrical circuits and is the basis for many important technologies, including electronics, power generation, and transmission.
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Water is flowing in a circular pipe varying cross-sectional area, and at all points, the water completely fills the pipe.a) At one point in the pipe the radius is 0.150 m. What is the speed of the water at this point if the water is flowing into this pipe at a steady rate of 1.20 m3/s?b) At a second point in the pipe the water speed is 2.90 m/s. What is the radius of the pipe at this point?
The speed of water at the point with a radius of 0.150 m is 16.97 m/s while the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.
a) To find the speed of the water at a point of a circular pipe where the radius is 0.150 m if the water is flowing into this pipe at a steady rate of 1.20 m³/s, we'll use the equation;
Q = A₁V₁ = A₂V₂ Where Q = Flow rate (m³/s)A₁ = Cross-sectional area at one point (m²)V₁ = Velocity of water at one point (m/s)A₂ = Cross-sectional area at a second point (m²)V₂ = Velocity of water at the second point (m/s)At one point in the pipe, the radius is 0.150 m.Therefore, the cross-sectional area, A₁ is given by:
A₁ = πr₁² = π (0.150 m)² = 0.0707 m²Given that the water is flowing into the pipe at a steady rate of 1.20 m³/s, we can write;Q = A₁V₁1.20 m³/s = 0.0707 m² V₁V₁ = 1.20/0.0707V₁ = 16.97 m/s.Therefore, the speed of water at the point with a radius of 0.150 m is 16.97 m/s.
b) To find the radius of the pipe at a point where the water speed is 2.90 m/s, we'll use the same equation as in part (a);Q = A₁V₁ = A₂V₂At a second point in the pipe, the water speed is 2.90 m/s.Given that the water completely fills the pipe, we know that the volume flow rate, Q will remain constant at 1.20 m³/s.So, we have:
Q = A₁V₁ = A₂V₂We know that A₁ = πr₁²So, Q = πr₁²V₁Also, we know that A₂ = πr₂²So, Q = πr₂²V₂Since the volume flow rate is constant, we can equate both equations,πr₁²V₁ = πr₂²V₂Dividing both sides of the equation by π, we have;r₁²V₁ = r₂²V₂But we are interested in finding the radius of the pipe at the second point, r₂.So, we can express r₁ in terms of r₂ using the relationship between the cross-sectional areas;
A₁ = A₂r₁² = (A₂/A₁)²r₂²r₁ = r₂ (A₂/A₁)^(1/2).We know that A₁ = πr₁²We can find A₂ using the fact that the water completely fills the pipe;
A₁V₁ = A₂V₂πr₁²V₁ = A₂V₂π(0.150 m)²(16.97 m/s) = A₂(2.90 m/s)A₂ = π(0.150 m)²(16.97 m/s)/(2.90 m/s)A₂ = 0.0707 m²
So,r₂ = r₁(A₂/A₁)^(1/2)r₂ = 0.150 m × (0.0707 m²/π)/(0.0150 m²)^(1/2)r₂ = 0.0682 m. Therefore, the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.
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Weigh yourself on a weighing scale and the scale shows your normal weight. If you carefully stand on tiptoes, the scale reading will be
A) slightly more.
B) slightly less.
C) about half as much.
D) no different.
Answer: slightly more
Explanation:
it will be slightly less because it is partially taking away the heaviness from your natural weight. But it might also be slightly more because you are still kind of adding more weight onto the scale by putting more weight onto your toes making it heavier.
A light bulb used in a slide projector draws a current of 6 amperes when operating on 120 volts.. the power consumed by th light bulb in watts is? B.)a light bulb used in a slide projector draws a cuurent of 6 amperes when operating on 120 volts. the resistance of the light bulb in ohms is?
a..05
b.20
c.720
d.none
When a light bulb used in a slide projector draws a current of 6 amperes while operating on 120 volts, the power consumed by the light bulb in watts is 720, and the resistance of the light bulb in ohms is 20. Thus, the correct option is B.
Why the resistance of a light bulb is 20 ohms?When we know that the current drawn by a light bulb is 6 amperes and the voltage applied to it is 120 volts, we can easily calculate the resistance of the light bulb, as follows:
Resistance (R) = Voltage (V) / Current (I)
here, V = 120V and I = 6A
Therefore, the resistance of the light bulb is:
R = V/I = 120/6 = 20 Ohms
The formula used to calculate the power (P) consumed by a light bulb is:
P = V × I
Here, the voltage (V) applied to the light bulb is 120 volts and the current (I) drawn by the light bulb is 6 amperes. So, the power consumed by the light bulb is:
P = 120 × 6 = 720 watts
Hence, the power consumed by the light bulb in watts is 720, and the resistance of the light bulb is 20 ohms.
Therefore, the correct option is B.
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a wire of length l carrying a current i is placed in a magnetic field. the direction of the magnetic field is opposite the direction of the current. in this situation, the wire experiences a maximum force. select one: a. false b. true
The given statement is true because the angle between the magnetic field and the current is 180 degrees, which maximizes the sine function in the cross-product formula used to calculate the force.
The direction of the force is given by the right-hand rule, where the thumb points in the direction of the current, the fingers point in the direction of the magnetic field, and the palm points in the direction of the force.
When the direction of the magnetic field is opposite to the direction of the current, the wire experiences a maximum force.
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A 4.50kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80m/s )t +(0.61m/s^3 )t^3 What is the magnitude of the force F when 3.60s ?
The magnitude of the force F is 47 N when 3.60 s.
A 4.50 kg crate is suspended from the end of a short vertical rope of negligible mass.
An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by
y(t) = (2.80m/s )t +(0.61m/s^3 )t^3.
First, we will find the speed of the crate:
v(t) = dy(t)/dt => (v(t)) = 2.80 + 1.83t^2
We have to find the magnitude of the force F(t) when t = 3.60 s.
Since the acceleration due to gravity is 9.81 m/s^2 and
the net force on the crate is 0, the upward force applied F(t) is equal to the weight of the crate.
W = mg => F(t) = 4.50 kg x 9.81 m/s^2= 44.14 N.
Using the equation of motion:
y(t) = 0.5gt^2 + v(0)t + y(0)
where g is the acceleration due to gravity,
v(0) is the initial speed of the object, and
y(0) is the initial position of the object,
we find the value of y(3.60) = 47.25 m.
Substituting t = 3.60 s, we get:
47.25 = 0.5 x 9.81 x (3.60)^2 + (2.80)(3.60) + (0.61/3.60^2) x (3.60)^3
After solving for the above expression, we get the magnitude of the force F when 3.60 s as 47 N.
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The following arrangement, consisting of a massless plate supported by 3 pillars and holding a
10−kg
mass, is in static equilibrium. Calculate the normal force exerted by all three pillars. You can assume that, on each individual pillar, the net normal force is applied in the middle of the face touching the plate. All dimensions are from centerto-center of the objects.
The normal force exerted by all three pillars in the given arrangement is 32.7 N.
To calculate the normal force exerted by each pillar, we can first find the weight of the 10-kg mass:
w = mg = (10 kg)(9.8 m/s^2) = 98 N
Since the mass and plate are in static equilibrium, the net force acting on the mass and plate must be zero. Therefore, the sum of the normal forces exerted by each pillar must equal the weight of the mass:
F1 + F2 + F3 = w
We can also use the fact that the normal force is equal and opposite to the force exerted by the mass on the pillars:
F1 = -f, F2 = -f, F3 = -f
where f is the force exerted by the mass on each pillar.
Therefore, we can rewrite the equation as:
-f - f - f = -3f = -w
Solving for f, we get:
f = w/3 = 98 N / 3 ≈ 32.7 N
Therefore, the normal force exerted by each pillar is approximately 32.7 N.
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An object speed is increased by a factor of three. What does this do to its kinetic energy?a) the kinetic energy increases by a factor of threeb) the kinetic energy increases by a factor of twoc) the kinetic energy increases by more than a factor of threed) the kinetic energy cannot be determinede) the kinetic energy increases, but less than by a factor of twof) It does not affect the kinetic energy
c) the kinetic energy increases by more than a factor of three. A three-fold increase in an object's speed occurs. The kinetic energy rises by a factor of greater than three.
An object's kinetic energy (KE) is determined by the equation KE = 1/2mv2, where m is the object's mass and v is its velocity. An object's kinetic energy is multiplied by nine (32) when its velocity is raised by a factor of three. This is due to the fact that kinetic energy is inversely proportional to square of velocity, meaning that any change in velocity will have a bigger impact on kinetic energy.
It follows that if an object's speed is raised by a factor of three, its kinetic energy will also rise by a factor of three or more.
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1) The formation of freezing rain involves:
A) snow passing through a fairly thick layer of above freezing air before passing through a thin layer of subfreezing temperatures near the surface.
B) air temperatures decreasing uniformly with height, producing the cold conditions necessary for freezing rain formation.
C) air temperatures increasing uniformly with height, producing the cold conditions necessary for freezing rain formation.
D) snow passing through a fairly thin layer of above freezing air before passing through a thick layer of subfreezing
temperatures near the surface.
The diagram shows a homemade car being pushed with a force of 25 N.
Answer:
The speed of the car will increase.
Explanation:
From idea of momentum, force is directly proportional to velocity
[tex]{ \bf{f \: \alpha \: v}} \\ { \rm{f = kv}}[/tex]
Initially, f = 25N and v = 3 m/s
[tex]{ \rm{25 = k \times 3}} \\ \\ { \rm{k = \frac{25}{3} }}[/tex]
Lastly, f = 35
[tex]{ \rm{f = \frac{25}{3}v }} \\ \\ { \rm{35 = \frac{25}{3} \times v}} \\ \\ { \rm{v = \frac{3 \times 35}{25} }} \\ \\ { \rm{v = 4.2} }[/tex]
which unit is used to measure force?(1 point) responses watt watt kilogram kilogram newton newton joule
According to the International System of Units, Newton (N) is used to measure force. Thus Newton is the correct response.
A force is defined as a physical quantity that causes a change in the state of motion of a body.
Force is a vector quantity that has both magnitude and direction.
Newton is the unit of force in the International System of Units (SI).
One newton (N) is the amount of force required to give a mass of 1 kg and an acceleration of 1 m/s².
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a train moves from a train station at 30meter per second after 5 seconds its cover the distance of 100 m and the acceleration is ten meters per second square find the speed
The speed of a train that moves from a train station at 30 meters per second after 5 seconds and covers a distance of 100 m with an acceleration of ten meters per second square would be 80 m/s.
Speed of a trainWe can use the equation of motion to solve for the final velocity of the train:
v = u + at
where:
v = final velocityu = initial velocity = 30 m/sa = acceleration = 10 m/s^2t = time = 5 sSubstituting the values, we get:
v = 30 + 10(5)
v = 80 m/s
Therefore, the speed of the train after 5 seconds is 80 m/s.
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This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Pin A, which is attached to link AB, is constrained to move in the circular slot CD. At t=0, the pin starts from rest and moves so that its speed increases at a constant rate of 1.2 in/s2 D 3.5 in. А B Determine the magnitude of its total acceleration when t= 0. The magnitude of its total acceleration is in/s2
The magnitude of the total acceleration of the pin when t=0 is 1.2 in/s^2.
To explain further, the acceleration of the pin is the sum of two components: tangential acceleration and centripetal acceleration. The tangential acceleration is responsible for increasing the speed of the pin, and its magnitude is constant at 1.2 in/s^2.
The centripetal acceleration is due to the circular motion of the pin in the slot CD and is directed towards the center of the circle.
To find the magnitude of the total acceleration at t=0, we need to first find the magnitude of the tangential acceleration and the centripetal acceleration separately. We know that the tangential acceleration is 1.2 in/s^2, and we can use the formula for centripetal acceleration, a_c = v^2/r, where v is the velocity of the pin and r is the radius of the circle. At t=0, the velocity of the pin is zero, and the radius of the circle is 3.5 inches.
Therefore, the centripetal acceleration is also zero.
Since the centripetal acceleration is zero, the magnitude of the total acceleration is equal to the magnitude of the tangential acceleration, which is 1.2 in/s^2.
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A 12100 kg railroad car is coasting on a level, frictionless track at a speed of 19.0 m/s when a 4790 kg load is dropped onto it.
If the load is initially at rest, find the new speed of the car and the % change of the kinetic energy.
Hint 1: If the load is dropped into the car, it is like the car is "colliding� with a stationary load. If the load is stuck in the car, can they have different final velocities from one another?
The percent change in the kinetic energy of the system is [(0.5*(12100 + 4790)*1722) - 5.58 x 106] / (5.58 x 106) x 100% = 4.41%.
The 12100 kg railroad car is initially travelling at a speed of 19.0 m/s and has a kinetic energy of KE = 0.5*12100*1902 = 5.58 x 106 Joules. The 4790 kg load is dropped onto the car from rest, so its initial kinetic energy is 0.
When the load is dropped onto the car, the two objects collide and their velocities after the collision will be equal. Therefore, the final speed of both the railroad car and the load will be v = (12100*19 + 4790*0) / (12100 + 4790) = 17.2 m/s. The percent change in the kinetic energy of the system is [(0.5*(12100 + 4790)*1722) - 5.58 x 106] / (5.58 x 106) x 100% = 4.41%.
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How does changing the mass of the black hole affect its Schwarzschild radius? Specifically, if we double the mass of the black hole, which of the following statements is true?The volume from which light cannot escape will get bigger.The Schwarzschild radius of the black hole also doubles.If the mass of the black hole doubles, the Schwarzschild radius of the black hole also doubles.A small, stellar-mass black hole with three times the mass of the Sun (5.97×1030 kg5.97×1030 kg) is roughly 9 km across, about the size of a city of a few tens of thousands of people.The Schwarzschild radius depends on the mass, MBMMBM, of the black hole.
Option 2) is true. If the mass of the black hole doubles, then the Schwarzschild radius of the black hole also doubles.
The Schwarzschild radius of a black hole depends on its mass. When the mass of the black hole changes, its Schwarzschild radius is also affected. Specifically, if we double the mass of a black hole, the Schwarzschild radius of the black hole also doubles. The Schwarzschild radius is defined as the distance from the center of a black hole at which an object would need to be in order to escape its gravitational pull. It is represented by the equation Rs = 2GM/c² where Rs is the Schwarzschild radius, G is the gravitational constant, M is the mass of the black hole, and c is the speed of light.
The formula implies that Schwarzschild radius depends on the mass of the black hole.Therefore, if the mass of the black hole doubles, then the Schwarzschild radius of the black hole also doubles. Hence, option 2 is the correct statement.
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what is the relationship between the velocity of a fluid and the size of the sediment that the fluid carries?
The relationship between the velocity of a fluid and the size of the sediment that the fluid carries is directly proportional.
Higher velocity fluids are capable of carrying larger sediments while lower-velocity fluids are capable of carrying smaller sediments. This is due to the fact that higher-velocity fluids have greater kinetic energy, which allows them to overcome the gravitational forces that hold larger sediments in place.
A fluid is a substance that is able to flow and take on the shape of the container it is placed in, with the ability to deform under applied shear stress. Examples of fluids include liquids and gases. In contrast, solids maintain their shape and volume under applied stress.
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does air move from areas of high pressure to low pressure
Explanation: Gases move from high-pressure areas to low-pressure areas. And the bigger the difference between the pressures, the faster the air will move from the high to the low pressure.
a 37.5 kg box initially at rest is pushed 5.40 m along a rough, horizontal floor with a constant applied horizontal force of 150 n. if the coefficient of friction between box and floor is 0.300, find the following. a. the work done by the applied forceb. the increase in internal energy in the box-floor system due to frictionc. the work done by the normal forced. the work done by the gravitational forcee. the change in kinetic energy of the boxf. the final speed of the box
A 37.5 kg box initially at rest is pushed 5.40 m along a rough, horizontal floor with a constant applied horizontal force of 150 N. If the coefficient of friction between box and floor is 0.300, the following can be found:
a. The work done by the applied force is equal to the product of the force and the displacement
Work = Fd = (150N)(5.40m) = 810 J.
b. The increase in internal energy in the box-floor system due to friction is equal to the work done by the friction force, Work done by friction = Frd = (μN)(5.40m) = (0.3)(37.5kg*9.8m/s²)(5.40m) = 546.1 J.
c. The work done by the normal force is zero because the normal force does not cause displacement,
Work done by normal force = 0.
d. The work done by the gravitational force is equal to the product of the gravitational force and the displacement, Work done by gravitational force = Fd = (mg)(5.40m) = (37.5kg*9.8m/s²)(5.40m) = 1817.9 J.
e. The change in kinetic energy of the box is equal to the work done by the applied force minus the work done by the friction and gravitational forces,
Change in kinetic energy = Work done by applied force - Work done by friction - Work done by gravitational force = 810 J - 546.1 J - 1817.9 J = -655 J.
f. The final speed of the box can be calculated using the equation
KE = 1/2mv2,
thus the final speed of the box is
v = sqrt(2KE/m) = sqrt(2(-655 J)/(37.5 kg)) = 1.62 m/s.
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use kinetic theory to explain what causes gas pressure
The kinetic molecular theory can be used to explain each of the experimentally determined gas laws. The pressure of a gas results from collisions between the gas particles and the walls of the container. Each time a gas particle hits the wall, it exerts a force on the wall.
A scientist is studying an organism that is similar to early life on Earth. The scientist observes structures form in the organism that appear as oily spheres with an inner fluid. Of which type of macromolecule is the sphere made? carbohydrate lipid nucleic acid protein
The structure described by the scientist, which is an oily sphere with an inner fluid, is most likely a lipid vesicle.
Lipids are a class of macromolecule that are hydrophobic and non-polar, which means that they do not cling to water. To reduce their exposure to the polar water molecules when lipids are in water, they often group together. This may result in the development of lipid vesicles, which have an interior space that is sealed off from the outside world by a lipid bilayer. Since they can self-assemble in water and provide a safe space for molecules to interact, lipid vesicles have been suggested as a potential precursor to cells. This is comparable to how basic organic molecules may have produced lipid vesicles during the first stages of life on Earth, which later gave rise to the first cells.
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X-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 7.84 ms. If the blob were in a circular orbit about a black hole whose mass is 13.5 times the mass of the Sun, what is the orbit radius? The value of the gravitational constant is 6.67259×10−11N⋅m2/kg2 and the mass of the Sun is 1.991×1030 kg. Answer in units of km.
The orbit radius of the blob in a circular orbit about the black hole is approximately 33,288 km.
The orbit radius of a blob in a circular orbit about a black hole whose mass is 13.5 times the mass of the Sun can be calculated using the formula:
r = (GMT²/4π²)1/3, where G is the gravitational constant, M is the mass of the black hole, and T is the period of the orbit.
X-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 7.84 ms. Therefore,
T = 7.84 × 10⁻³ seconds
M = 13.5
Mʘ = 13.5 × 1.991 × 10³⁰ kg = 2.68585 × 10³¹ kgG = 6.67259 × 10⁻¹¹ N m²/kg²
Now, substituting the given values in the formula:
r = [(6.67259 × 10⁻¹¹ × 2.68585 × 10³¹ × (7.84 × 10⁻³)²) / (4π²)]1/3r = 33,288,375 meters ≈ 33,288 km
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