Answer:
Spiral, Elliptical, and Irregular
Explanation:
1.Spiral galaxies feature spiralling arms that stretch from their centres, giving them a pinwheel look. The majority of the mass in these galaxies is located in a bulge, which is a denser, spherically shaped region. The remaining mass is distributed throughout the arms, which are spirals of stars and dust that spiral relatively in the same direction.
2.Elliptical galaxies have an elliptical shape, lack arm-like features, and seem to contain little interstellar materials, such as black dust clouds. These galaxies also seem to have low rates of star formation and, as a result, low levels of heavy elements produced by stars. Elliptical galaxies could form as a result of two or more galaxies joining together.
3.The form of irregular galaxies is completely lacking. They lack spiral arms and a galactic nuclear bulge.
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Fluid flows through a 0.22m diameter pipe at a velocity of 2.8m/s.
Calculate the cross sectional area of the pipe.
m2
Calculate the Volumetric Flow Rate (Q) of the fluid.
m3/s
How long would it take for 3.3m3 of fluid to flow through the pipe?
Answer:
9.22 m=s
Explanation:
m=s×v
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Horsepower is defined as ______.
A. power applied over the distance of an English horserace track
B. the mechanical advantage gained by using a typical horse
C. the mass of a horse divided by the time it takes the horse to run 100 meters
D. a specific amount of power (550 ft-lb/s)
The definition of horsepower is . A. the amount of force used to cover the length of an English associated activities track B. the practical benefit of utilizing a standard horse
Who developed the horsepower?
James Watt, an engineer, is credited with creating the term horsepower. According to legend, Watt wanted to find a method to describe the power that one of those animals could produce while he was dealing with ponies in a coal mine. An engine is connected to a dynamometer in order to determine its horsepower.
What is the power of a horse?
The most prevalent unit of power is horsepower , which measures how quickly work is completed. According to the British Imperial Units, a horsepower is equivalent to 33,000 walking of work per minute, or the force required to elevate a mass of 3 million pounds one foot in a minute.
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earth moves about 30 km/s relative to the sun. when you jump upward in front of a wall, the wall doesn't slam into you at 30 km/s because the wall
The Earth's motion is unrelated to your motion, so the wall doesn't slam into you at 30 km/s.
What is Motion?Motion is the change in an object's location with regard to its surroundings over a specific amount of time.
While the earth moves relative to the sun at a speed of around 30 km/s, this velocity has no bearing on how things move on the surface of the planet. Because the motion of things on the surface of the earth is governed by the gravity of the earth rather than its velocity relative to the sun, when you jump upward in front of a wall, the wall does not slam into you at a speed of 30 km/s.
The uniform gravitational field produced by the earth's gravity causes things on its surface to accelerate steadily in the direction of the planet's center. Your muscles generate an upward push that causes you to jump upward, leaving the ground.
Similar gravitational effects are also felt by the wall, which maintains its equilibrium with respect to the surface of the earth. Since no outside force is acting on the wall to cause it to shift position when you jump upward, the wall does not move towards you.
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Choose one human-made object in motion on Earth. Draw a picture of the object and add arrows showing the forces acting on the object while it is in motion. Be sure to show the overall direction in which the object is moving by scaling your arrows.
Be sure to label the forces.
You need to include at least 2 forces for full credit.
The arrow pointing downward is gravitational force, F(g), The left arrow is the applied force, F(t), The right arrow is frictional force, F(f) and the arrow pointing upward is the net force, F(n).
What is a force?A force is a push or pull upon an object resulting from the object's interaction with another object. Forces can cause an object to accelerate or change its direction of motion.
Forces can be described by their magnitude and direction. In physics, forces can be caused by many interactions, including gravity, electromagnetic forces, and the strong and weak nuclear forces.
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A mass on a string of unknown length oscillates as a pendulum with a period of 3.3 s. What is the period if a. The mass is doubled?b. The string length is doubled?c. The string length is halved?d. The amplitude is doubled?
a. The mass is doubled = 4 s
b. The string length is doubled= 5.66 s
c. The string length is halved=2.83 s
d.The amplitude is doubled= 4 s
How to calculate period of pendulum(a) The period of the pendulum is independent of the mass, therefore, the period when the mass is doubled is
T= To =4.00s
(b) Let I be the new length of the pendulum and Lo be the original length of the pendulum. The period of the pendulum if the string length is doubled is found by substituting 2Lo for L in Equation (*):
2π=√2Lo/g
√2=(2π(√Lo/g)
where the term in parenthesis is the original period of the pendulum To:
T=√√2To
T=√2.(4seconds)
T=5.66 s
c) The period of the pendulum if the string length is halved is found by substituting Lo/2 for L in Equation (*):
T=2π(√Lo/2 :g)
T= 1/√2 (2π√(Lo/g)
where the term in parenthesis is the original period of the pendulum To:
T =To /√2
T=4.00s/√2
T=2.83 s
d.) The period of the pendulum is independent of the amplitude, therefore, the period when the amplitude is halved is
T= To 4.00s
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The following circuit diagram is partially incomplete.
The current in Lamp 1 is 0.4 A. The potential difference across Lamp 1 is 8.0 V.
Calculate the resistance of Lamp 1.
Group of answer choices
A) 30 ohms
B) 20 ohms
C) 2 ohms
D) 3 ohms
Answer:
20Ω
Explanation:
we are here given that,
Potential difference= 8Vcurrent= 0.4 A Resistance= ?From Ohm's law ,
[tex]\implies V = iR \\[/tex]
where,
V is potential differencei is currentR is resistanceon substituting the respective values, we have,
[tex]\implies 8V = 0.4A \times R\\[/tex]
[tex]\implies R =\dfrac{8V}{0.4A} \\[/tex]
[tex]\implies \underline{\underline{ R = 20\Omega}}\\[/tex]
and we are done!
When connected to a 12V battery, the current in a car headlight is 4.0 A.
When the lamp is lit, what is its resistance?
A) 3.0 ohms
B) 16 ohms
C) 48 ohms
D) 0.33 ohms
Answer:
A) 3.0 ohms
Explanation:
R=V/I=12V/4.0A=3.0 ohms
he circular orbits of satellites 1 and 2 coincide. Satellite 2 has twice the mass of satellite 1. Compare their accelerations. A) 1's acceleration is half as much. B) 1's acceleration is the same as 2's C) 1's acceleration is twice as much as 2's D) It depends on the periods of their orbits.
The correct answer is (a): satellite 1's acceleration is half as much as satellite 2's.
This is due to the fact that the acceleration of two satellites in a circular orbit depends on their masses and orbital periods. Satellite 2 accelerates more quickly because it has twice as much mass as satellite 1. As a result, satellite 1 accelerates at a rate that is half that of satellite 2.
This results in a divergence in the orbits of the two spacecraft since satellite 1's speed will rise more slowly than satellite 2's. Over time, this difference will become more obvious, with satellite 1's orbit straying from a circular form more so than satellite 2's.
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when monkey mo is suspended at rest by holding a rope with one hand and the side of his cage with the other, all the force vectors that act on him
Monkey Mo is in a static equilibrium because of the balance of all these forces acting on him.
What is force?The interaction between two things or the pressure applied to alter an object's motion is described by the physical quantity known as force. An object can change direction, accelerate, decelerate, or continue in motion as a result of force.
The forces at work on Monkey Mo when he is hung at rest while grasping a rope in one hand and the side of his cage in the other are as follows:
Gravitational force, which is vertically downward operating force of attraction between Monkey Mo and the Earth.
The tension force, which acts upward along the length of the rope, is the force the rope applies to Monkey Mo.
Normal force, which force balances the gravitational pull by operating perpendicular to the surface of the cage and being applied to Monkey Mo by the side of the cage.
Frictional force prevents Monkey Mo from falling out of the cage by acting between his hand and the side of the cage.
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Transverse waves on a string have wave speed 12m/s, amplitude 0.05m and wavelength 0.4m. The waves travel in the positive x direction, and at t=0 the x = 0 end of the string has zero displacement and is moving upward.
(a) Find the frequency, period and wave number of these waves
(b)Write a wave function describing the wave
(c) Find the transverse displacement of a wave at x=0.25 at time t -0.15sec
(d) How much time must elapse from the instant in part "(c)" until the point at x = 0.25m has zero displacement
An automobile tire has a volume of 0.0185 m³. At a temperature of 289 K the absolute pressure in the tire is 217 kPa. How many moles of air must be pumped into the tire to increase its pressure to 272 kPa, given that the temperature and volume of the tire remain constant?
0.0018 moles of air must be pumped into the tire to increase its pressure to 272 kPa.
How to find the number of moles of airWe can use the ideal gas law to find the number of moles of air in the tire.
The ideal gas law states that PV = nRT,
where
P is pressure,
V is volume,
n is the number of moles of gas,
R is the ideal gas constant, and
T is temperature.
The initial number of moles of air in the tire:
n1 = (P1 * V) / (R * T)
= (217 kPa * 0.0185 m³) / (8.31 J/mol * K * 289 K)
= 0.00625 moles
Next, we can find the final number of moles of air in the tire, assuming the temperature and volume remain constant:
n2 = (P2 * V) / (R * T)
= (272 kPa * 0.0185 m³) / (8.31 J/mol * K * 289 K)
= 0.00805 moles
Finally, we can find the difference between the final and initial number of moles, which represents the number of moles of air that must be pumped into the tire:
n2 - n1 = 0.00805 moles - 0.00625 moles = 0.0018 moles.
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Which is NOT a function of the skeletal system?
Answer:produces red and white blood cells
Explanation: the bone marrow helps with the blood but cannot produce red and white blood cells
Answer:
"prevents movement of limbs and digits" is NOT a function of the skeletal system.
The equivalent resistance of 2.0 Ω, 3.0 Ω, 5.0 Ω, and 6.0 Ω resistors connected in parallel is ________.
A) 1.2 Ω
B) 7.2 Ω
C) 0.83 Ω
D) 9.3 Ω
Answer:
C) 0.83 Ω
Explanation:
A material has a stress–strain relationship that can be approximated by the equation
ε=0.3x10-16xσ3
where the stress is in psi. Find the secant modulus and the tangent modulus for the stress level of 50,000 psi.
The secant modulus and the tangent modulus for the stress level of 50,000 psi are both 4.5x10-9 psi-1.
What is secant modulus?Secant modulus is a measure of a material's ability to resist deformation under stress. It is the ratio of stress to strain in a material and is equal to the slope of a line drawn between two points on a stress-strain curve.
The secant modulus is the slope of the line connecting two points on the stress-strain curve. In this case, the two points we need to consider are (50,000 psi, ε) and (0 psi, 0).
Therefore, the secant modulus is given by:
secant modulus = (ε - 0) / (50,000 psi - 0 psi)
Plugging in the equation for ε, we get:
secant modulus = (0.3x10-16x50,0003 - 0) / (50,000 psi - 0 psi)
secant modulus = 4.5x10-9 psi-1
The tangent modulus is the slope of the tangent line to the stress-strain curve at a given point. In this case, the point is (50,000 psi, ε).
Therefore, the tangent modulus is given by:
tangent modulus = dε/dσ
Plugging in the equation for ε, we get:
tangent modulus = (3x10-16x50,0002) / (1 psi)
tangent modulus = 4.5x10-9 psi-1
Therefore, the secant modulus and the tangent modulus for the stress level of 50,000 psi are both 4.5x10-9 psi-1.
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A sphere and a cylinder of equal mass and radius are simultaneously released from rest on the same inclined plane and roll without sliding down the incline. Then: A. the sphere reaches the bottom first because it has the greater inertia B. the cylinder reaches the bottom first because it picks up more rotational energy C. the sphere reaches the bottom first because it picks up more rotational energy D. they reach the bottom together E. none of the above are true
The correct option for the problem is option D. they reach the bottom together. Their gravitational potential energy is converted to kinetic energy, total energy will be conserved for the system and we can safely say they will reach the bottom together without considering their shape.
Here sphere and cylinder both have equal mass and equal radius and it is given that both are simultaneously released also.
When two objects with equal mass and equal radius like these two, the time taken for them to roll down and reach the bottom will be their gravitational potential energy.
Here what happens is that their gravitational potential energy is converted to kinetic-energy and both of them rolling down the incline.
Total energy will be conserved for the system and we can safely say they will reach the bottom together without considering their shape.
Inertia and rotational energy are also there in this context for their motion but they do not take part any role for the time taken to slide down.
Instead, they affect the stability and trajectory of the objects as they roll down the incline.
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Find the work W done if a constant force of 85 lb is used to pull a cart a distance of 220 ft. W=_____ft-lb
If a constant force of 85 lb is used to pull a cart a distance of 220 ft. Then the work done W is 18700 ft-lb.
Work is the amount of energy transferred when a force is applied over a distance. In this case, a constant force of 85 lb is used to pull a cart a distance of 220 ft. To find the work done, we can use the formula W = Fd, where F is the force and d is the distance. Plugging in the values, we get W = 85 x 220 = 18700 ft-lb. This is the amount of work done by the force to move the cart a distance of 220 ft.
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Two weights are connected by a massless wire and pulled upward with a constantspeed of 1.50 m/s by a vertical pull P. The tension in the wire is T(see figure). Whichone of the following relationships between Tand Pmust be true?
A)T
B)T=P
C)P+T=125N
D)P=T+25N
The correct relationship between T and P is represented by option B)
T = P.
The net force operating on the system is represented by the tension in the wire T, which is equal to the total of the forces acting on each weight. Because only the gravitational force and the tension force are operating on each weight in this situation, the net force exerted on each weight is equal to the difference between the tension and the weight. T must therefore equal the sum of the gravitational pull P and the forces acting on each weight.
Since m represents the mass of each weight and g is the acceleration brought on by gravity, the equation for the relationship between T and P must be T = P + m * g.
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A positive charge moves in the direction of an electric field. Which of the following statements are true?
Check all that apply.
A positive charge moves in the direction of an electric field. Which of the following statements are true?Check all that apply.
The potential energy associated with the charge decreases.
The electric field does not do any work on the charge.
The electric field does positive work on the charge.
The amount of work done on the charge cannot be determined without additional information.
The potential energy associated with the charge increases.
The electric field does negative work on the charge.
The first and third statements for the positive charge which moves in the electric field are correct.
When placed in an electric field, the positive charge typically flows in the direction of the electric field. The work done by the electric field in this situation is positive because both the electric field vector and the displacement vector point in the same direction and there is no angle between them.
The potential will also decrease in the direction of the electric field. This is because the work done in the order of the electric field is called the potential. electric potential is the amount of work required to move a unit charge from a reference point against an electric field to a particular point
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When a stone is projected, its horizontal range is 24m and greatest height 6m. Find its velocity of projection.
Answer:
15.33 m/s
Explanation:
We are here given that ,
Range = 24m (R)Maximum height= 6m (h)velocity of projection= ? (u)As we know that,
[tex]\longrightarrow R = \dfrac{u^2\sin2\theta}{g} \dots (1)\\[/tex]
And ,
[tex]\longrightarrow h =\dfrac{u^2\sin^2\theta}{2g} \dots (2)\\[/tex]
Divide equation 1 and 2 ,
[tex]\longrightarrow \dfrac{R}{h}=\dfrac{4 \cos\theta}{\sin\theta} \\[/tex]
[tex]\longrightarrow \dfrac{24m}{6m}=4 \cot\theta \\[/tex]
[tex]\longrightarrow \cot\theta = 1 \\[/tex]
[tex]\longrightarrow \theta = \cot^{-1}(1) \\[/tex]
[tex]\longrightarrow \underline{\underline{\theta = 45^{\circ}}}\\[/tex]
Now we may substitute this value in equation 1 as ,
[tex]\longrightarrow 24 =\dfrac{ u^2\sin(2\times 45^{\circ} )}{g} \\[/tex]
[tex]\longrightarrow 24g = u^2\sin90^\circ \\[/tex]
[tex]\longrightarrow u^2 = 24 \times 9.8 \\[/tex]
[tex]\longrightarrow u =\sqrt{235.2} m/s \\[/tex]
[tex]\longrightarrow \underline{\underline{ u \approx 15.33 \ m/s }}\\[/tex]
and we are done!
Can someone answer this pls ?
It is discovered that the object, which weighs 60g and has a volume of 50cm³, has a density of 1.2 g/cm³.
Does density vary as a result of pressure?Under pressure, a liquid becomes denser. In other words, as an excessive force is given to a liquid, volume decreases and density increases, and as molecules get closer to one another, interparticle spacing decreases.
What does mass vs. density mean?A substance or object's mass is a measure of how much matter it contains. Volume (the amount of space an object or substance occupies) to mass (the amount of material present) ratio is known as density (its mass).
According to the given information:
Mass= 60g
Volume = 50cm³
Density = mass / volume
=> 60/50 =>1.2 g/cm³
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In the following problems you will be asked to calculate the net gravitational force acting on the Moon. To do so, please use the following variables. Mass msun = m1 mEarth = m2 mMoon = m3Initial Position Sun =< 0,0,0 > PEarth =< 1,0,0 > Moon =< L, h,0 > (a) Calculate the gravitational force on the Moon due to the Earth. "G" is the universal gravitational constant. (b) Calculate the gravitational force on the Moon due to the Sun. (c) Determine the net gravitational force on the Moon. < >
The net gravitational force is terms of L and h and is equal to[tex]f_{net} = G m_3*(m_2 / ((L - 1)^2 + h^2) + m_1 / (L^2 + h^2))[/tex]
(a) The gravitational force on the Moon due to the Earth can be calculated using the equation:
f_gravity = G * (m2 * m3) / d^2
where d is the distance between the Earth and the Moon. The distance between the Earth and the Moon can be calculated using the Pythagorean theorem:
d = sqrt((L - 1)^2 + h^2)
Plugging in the values, we get:
[tex]f_{gravity} = G * (m_{Earth} * m_{Moon}) / d^2 = G * (m_{2} * m_{3}) / d^2 \\\\= G * (m_{2} * m_{3}) / (sqrt((L - 1)^2 + h^2))^2[/tex]
(b) The gravitational force on the Moon due to the Sun can be calculated using the same equation as above:
f_gravity = G * (m1 * m3) / d^2
where d is the distance between the Sun and the Moon. The distance between the Sun and the Moon can be calculated using the Pythagorean theorem:
d = sqrt(L^2 + h^2)
Plugging in the values, we get:
[tex]f_{gravity} = G * (m_{Sun} * m_{Moon}) / d^2 \\= G * (m_{1} * m_{3}) / d^2 = G * (m_{1} * m_{3}) / (sqrt(L^2 + h^2))^2[/tex]
(c) The net gravitational force on the Moon is the sum of the gravitational forces due to the Earth and the Sun:
f_net = f_gravity_Earth + f_gravity_Sun
Substituting the values of f_gravity_Earth and f_gravity_Sun, we get:
[tex]f_{net} = G * (m_2 * m_3) / (sqrt((L - 1)^2 + h^2))^2 + G * (m_1 * m_3) / (sqrt(L^2 + h^2))^2[/tex]
[tex]f_{net} = G * (m_2 * m_3) / ((L - 1)^2 + h^2) + G * (m_1 * m_3) / (L^2 + h^2)[/tex]
[tex]f_{net} = G m_3*(m_2 / ((L - 1)^2 + h^2) + m_1 / (L^2 + h^2))[/tex]
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a satellite of mass m moves in a circular orbit of radius r at a constant speed v. which of the following must be true? i. the net force on the satellite is equal to mv2 /r and is directed toward the center of the orbit. ii. the net work done on the satellite by gravity in one revolution is zero. iii. ithe angular momentum of the satellite is a constant.
(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III
The answer is (E) I, II, and III. When a satellite of mass m moves in a circular orbit of radius r at a constant speed v, all three statements are true.
i. The net force on the satellite is equal to mv^2 / r and is directed toward the center of the orbit. This force is known as the centripetal force, and it keeps the satellite moving in a circular path.
ii. The net work done on the satellite by gravity in one revolution is zero. Work is defined as the dot product of force and displacement, and the displacement of the satellite in a circular path is zero, so the net work done on it is also zero.
iii. The angular momentum of the satellite is a constant. Angular momentum is a measure of an object's rotational motion and is conserved if there are no external torques acting on the system. In this case, the satellite is in a stable orbit and there are no external torques acting on it, so its angular momentum is conserved and remains constant.
Therefore, all three statements (I, II, and III) must be true.
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at a distance r1 from a point charge, the magnitude of the electric field created by the charge is 397 n/c. at a distance r2 from the charge, the field has a magnitude of 145 n/c. find the ratio r2/r1.
The distance ratio r2/r1 is approximately 0.605.
How to calculate the ratio?The point charge is given by
E = k*q/r^2
where E is the electric field, k is the Coulomb constant ([tex]k = 8.99 * 10^9 Nm^2/C^2[/tex]), q is the charge of the point charge, and r is the distance between the point charge and the location. The electric field is measured.
Using this equation, we can find the distance ratio r2/r1.
[tex]E1 = k*q/r1^2\\E2 = k*q/r2^2[/tex]
Dividing these equations gives:
[tex]E2/E1 = (kq/r2^2) / (kq/r1^2)\\E2/E1 = (r1/r2)^2[/tex]
Solving for r2/r1 gives:
r2/r1 = √(E2/E1)
Inserting the given value will result in:
r2/r1 = √(145n/C ÷ 397n/C)
r2/r1 = √(0.365)
r2/r1 = 0.605
Therefore, the distance ratio r2/r1 is approximately 0.605.
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the gravitational force exerted by the planet earth on a unit mass at a distance from the center of the planet is
The gravitational force put by the planet earth on a unit mass at a distance from the center of the planet is "the strength of the gravitational field at that location".
It is directly proportional to the mass of the planet and inversely proportional to the square of the distance between the unit mass and the center of the planet. The formula for gravitational force is
F = G * (m1 * m2) / r^2,
where
G is the gravitational constant,m1 and m2 are the masses of the planet and the unit mass, and r is the distance between the two.The strength of the gravitational field decreases with increasing distance from the planet.
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The condition for the photodynamic effect to occur is that the pigment must have
a. Phosphorescence
b. Fluorescence
c. The absorption spectrum is the same as the absorption spectrum of the substrate
The condition for the photodynamic effect to occur is that the pigment must have the absorption spectrum is the same as the absorption spectrum of the substrate.
What is photodynamic treatment ?In order to induce cell death (phototoxicity), photodynamic treatment (PDT), a type of phototherapy, uses light, a photosensitizing agent, and molecular oxygen.
PDT is frequently used to treat acne. It is clinically used to treat a variety of medical problems, such as wet age-related macular degeneration, psoriasis, and atherosclerosis. It has also showed some promise in the treatment of viral diseases like herpes. Malignant malignancies of the head and neck, bladder, lung, and specific skin types are also treated by it.
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Energy Essay Question 1
You do work on something when you lift it against gravity. How does this work relate to gravitational potential
energy? If the lifted object is released, what becomes of this energy? Be sure to define all terms that you use.
Posted Mon Jan 30, 2023 at 1:16 pm
Criteria
Grading Scale
DS Immersive Reader Grade: N/A O
As the object is raised, the gravitational potential energy of the system grows. When you drop something, the potential energy of the object is transformed into kinetic energy, which is the energy of the moving object, when it strikes the ground.
What is kinetic and potential energy?
Potential energy is the energy stored in an object or system due to its position or arrangement of elements. It is unaffected by the environment outside the installation or system, such as air or altitude. Conversely, kinetic energy refers to the energy of moving particles in a system or object.
An object's kinetic energy, unlike potential energy, relates to other stationary and moving objects in its immediate vicinity. Thus, the lifted object is in kinetic motion when released.
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If there were no such thing as friction, what would happen to the motion of the soccer ball? (Please answer quickly, it's due today)
**a person kicking a soccer ball**
The soccer ball would gradually slow down.
The soccer ball would travel only a short distance.
The soccer ball would keep traveling until it hit something else.
The soccer ball would stop instantly.
Answer:
The soccer ball would keep traveling until it hit something else.
Explanation:
Friction acts as a force that opposes motion and slows objects down. When friction is absent, there is no force slowing the soccer ball down. So, without friction, the soccer ball would continue to move in a straight line at a constant velocity (assuming no other forces are acting on the ball) until it encounters an obstacle such as a wall or the ground, at which point it would stop.
An initially motionless test car is accelerated uniformly to 145km/h in 8.28s before striking a simulated deer. The car is in contact with the faux fawn for .815s, after which the car is measured to be traveling at 82.0km/h. What is the magnitude of the acceleration of the car before the collision? What is the magnitude of acceleration of the car during the collision? What is the magnitude of acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
The magnitude of acceleration of the car during the entire test is 4.83 m/s^2.
To calculate the magnitude of acceleration of the car before the collision, we can use the formula for uniform acceleration:
a = (v_f - v_i) / t
where a is acceleration, v_f is final velocity, v_i is initial velocity, and t is time.
v_i = 0 (the car is initially at rest), v_f = 145 km/h = 40 m/s, t = 8.28 s
a = (40 - 0) / 8.28 = 4.83 m/s^2
To calculate the magnitude of acceleration of the car during the collision, we can use the formula for average acceleration:
a = (v_f - v_i) / (2t)
where v_f is the final velocity after the collision (82 km/h = 22.8 m/s), v_i is the initial velocity before the collision (145 km/h = 40 m/s), and t is the time of the collision (0.815 s).
a = (22.8 - 40) / (2 * 0.815) = -19.86 m/s^2
Finally, to find the magnitude of acceleration of the car during the entire test, we have to integrate the acceleration over the time interval from t = 0 to t = 8.28 s. The area under the acceleration versus time graph represents the velocity of the car. By finding the velocity at t = 8.28 s, we can find the acceleration required to get from rest to that velocity.
v = a * t = 4.83 * 8.28 = 40 m/s
a = v / t = 40 / 8.28 = 4.83 m/s^2
So the magnitude of acceleration of the car during the entire test is 4.83 m/s^2.
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A simple circuit exists with one light bulb. A resistor of equal resistance to the bulb gets placed in the circuit in parallel. Which of the following would occur?
A. The current would increase.
B. The total voltage would decrease.
C. The current would stay the same.
D. The total resistance would remain the same.
[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
If we add a new residence in the circuit in parallel to the bulb of same resistance, the resultant resistance will get half the value of initial resistance [ when only bulb was connected in the circuit ], the current will increase as resultant resistance is half the original. and same current will flow in both the parallel wires.
So, by above information :
A.) Current would increase ? [ Yes ], As by ohms law, V = IR
and, if R is halfed, to maintain the same potential difference, current in the circuit will increase.
B.) Total voltage will decrease ? [ No ]... voltage will remain constant
C.) The current would stay the same ? [ yes ], it would increase...
D.) Total resistance would remain same ? [ No ], it will decrease...
A star of mass 5 × 10^30 kg is located at ‹ 8 × 10^12, 5 × 10^12, 0 › m. A planet of mass 6 × 10^24 kg is located at ‹ 5 × 10^12, 8 × 10^12, 0 › m and is moving with a velocity of ‹ 0.5 × 10^4, 1.5 × 10^4, 0 › m/s. During a time interval of 1×10^6 seconds, what is the change in the planet's velocity (Vf - Vi)?
The change in the velocity of the planet is 18.55 m/s.
What is the change in the planets velocity?
The gravitational force between the star and the planet, is calculated by applying Newton's third law of motion.
F = Gm₁m₂ /r²
where;
r is the distance between the star and the planet\m₁ is the mass of the starm₂ is the mass of the planetThe distance between the start and the planet is calculated as;
r = √ ( 8 x 10¹² - 5 x 10¹² )² + ( 5 x 10¹² - 8 x 10¹² )²
r = 4.24 x 10¹² m
F = ( 6.67 x 10⁻¹¹ x 5 x 10³⁰ x 6 x 10²⁴ ) / ( 4.24 x 10¹² )²
F = 1.11 x 10²⁰ N
The acceleration of the planet is calculated as;
a = F / m₂
a = ( 1.11 x 10²⁰ N ) / ( 6 x 10²⁴ kg )
a = 1.85 x 10⁻⁵ m/s²
The change in the velocity of the planet;
Δv = at
Δv = ( 1.85 x 10⁻⁵ x 1 x 10⁶ s )
Δv = 18.55 m/s
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