The total distance of the student is 37 m. The total displacement of the student is 7 m south.
What is distance?
The distance of an object is the total space covered by the object. It is scalar quantity because it has only magnitude with no direction.
The total distance covered by the student is calculated as follows;
distance = 15 m + 7 m + 15 m
distance = 37 m
What is displacement?The displacement of an object is the change in the position of the object. Displacement is vector quantity because it has both magnitude and direction.
The displacement of the student is calculated as follows;
displacement = space between the 15 m west and 15 meter east = 7 m south.
Thus, the total distance of the student is 37 m. The total displacement of the student is 7 m south.
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The Earth’s radius is 6.4 x 10^6 m and 1kg = 2.2lb. What is the force that the Earth exerts on you?
Using what you found out in question 1, what is the force that you exert on the Earth?
What would the Sun’s force be on the Earth, if our planet were twice as far from the Sun as it is now?
How does that force (from number 3) compare to the force from the Sun at our present location?
The force the earth exerts on a person that is 1 kg is 9.8 N.
The force the person exert on earth is 9.8 N.
If our planet were twice as far from the Sun as it is now, the Sun’s force on the Earth will be quarter the initial value.
Force the Earth exerts on you
The force the earth exerts on you is calculated as follows;
F = mg
where;
g is the acceleration due to gravity of EarthF = 1 kg x 9.8 m/s²
F = 9.8 N
Thus, the force the earth exerts on an object that is 1 kg is 9.8 N.
Also, based on Newton's third law of motion, the force the exerts on you is equal to the force you exert on Earth.
Force between the Earth and SunF = GmM/R²
where;
m is mass of EarthM is mass of the sunG is gravitation constantR is the distance between the sun and the earthF = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1.98 x 10³⁰) / (150.23 x 10⁹)²
F = 3.5 x 10²² N
When our planet is twice as farF = GmM/(2R)²
F = GmM/(4R²)
F = ¹/₄GmM/R²
Thus, if our planet were twice as far from the Sun as it is now, the Sun’s force on the Earth will be quarter the initial value.
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100 points,
The National Park Service sometimes creates controlled burns to mitigate wildfires. How does a controlled burn limit the spread of wildfires?(1 point)
A controlled burn is used to make a path that helps a spreading wildfire arrive at a source of water.
A controlled burn temporarily shuts down parks so people won’t start campfires.
A controlled burn helps to remove plants and vegetation around buildings so a wildfire won’t destroy them.
A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
What is a controlled burns?The term controlled burn refers to setting up an area in which the fire is controlled in order to avoid wild fires. These are deliberately set up in order to avoid the bush from burning down.
Let us recall that a wild fire is able to blaze across a large causing damage to a buildings as well as life and other properties in the way of the fire and could cause huge looses including loss of habitat.
Thus, the National Park Service sometimes creates controlled burns to mitigate wildfires because a controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.
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in europe, gasoline efficiency is measured in km/l . if your car's gas mileage is 36.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142- km trip in europe? use the following conversions: 1km
By converting the unit, the gas needed to complete a 142 km trip is 11.15 liters.
We need to know how to convert from mile and gallon to km and liter units to solve this problem. The conversions unit is
1 mile = 1.60934 km
1 gal = 4.54609 liter
From the question above, we know that:
gas mileage = 36 mi / gal
distance = 142 km
Convert the gas mileage to km/l
gas mileage = 36 mi / gal
gas mileage = 36 x 1,60934/4.54609 km/l
gas mileage = 12.74 km/l
Find the required gas
required gas = distance / gas mileage
required gas = 142 / 12.74
required gas = 11.15 liters
Hence, the gas needed to complete a 142 km trip is 11.15 liters.
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6 A gun fires a shell at an angle of elevation of 30⁰ with a velocity of 2 x 10³ ms. What are the horizontal and vertical components of the velocity? What is the range of the shell? How high will the hall rise?
1. The horizontal component of the velocity is 1.73×10³ m/s
2. The vertical component of the velocity is 1×10³ m/s
3. The range of the shell is 3.53×10⁵ m
4. The maximum height reached is 1.53×10⁵ m
1. How to determine the horizontal component of the velocity
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Horizontal velocity =?Horizontal velocity = u × Cosθ
Horizontal velocity = 2×10³ × Cos30
Horizontal velocity = 1.73×10³ m/s
2. How to determine the vertical component of the velocity
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Vertical velocity =?Vertical velocity = u × Cosθ
Vertical velocity = 2×10³ × Sin30
Vertical velocity = 1×10³ m/s
3. How to determin the range
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Acceleration due to gravity (g) = 9.8 m/s²Range (R) =?R = u²Sine(2θ) / g
R = (2×10³)² × Sine (2×30) / 9.8
R = 3.53×10⁵ m
4. How to determine the maximum height
Initial velocity (u) = 2×10³ m/sAngle of projection (θ) = 30 °Acceleration due to gravity (g) = 9.8 m/s²Maximum height (H) =?H = u²Sine²θ / 2g
H = [(2×10³)² × (Sine 30)²] / (2 × 9.8)
H = 1.53×10⁵ m
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a college student rests a backpack upon his shoulder. the pack is suspended motionless by one strap from one shoulder.
The student's shoulder supports the weight of the bag.
What is the free body diagram?Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.
A university student is carrying a backpack. One strap is hanging the rucksack immobile from one shoulder.
The weight of the backpack is balanced by the shoulder of the student.
The free-body diagram is attached below.
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What happens to the interference pattern that results from the diffraction of electrons passing through two closely spaced slits if the rate of electrons going through the slits is decreased to one electron per hour?.
The interference pattern remains the same if the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process.
What is Interference?This is referred to as a condition which occurs when two waves meet and combine while traveling along the same medium such as air, water etc.
The interference pattern is the same if the the rate of electrons going through the slits is decreased to one electron per hour during a diffraction process. The decrease in the rate of electrons doesn't affect the pattern of the waves being a constructive or destructive one.
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A string of length L , mass per unit length μ, and tension T is vibrating at its fundamental frequency. (iii) If the tension is doubled, with all other factors held constant, what is the effect on the fundamental frequency? Choose from the same possibilities as in part (i).
The fundamental frequency will observe the change by the multiple of [tex]\sqrt{2}[/tex]
The fundamental frequency of a standing wave which has
string of length L
mass per unit meu
has the tension T
is given by f = 1/2L[tex]\sqrt{T/meu}[/tex]
Now we are required to find if the tension is doubled, with all other factors held constant, what will be the effect on the fundamental frequency
According to the formula,
the fundamental frequency f is directly proportional to the factor of square root
Hence, when the tension of the string is doubled while other factors of the system remain constant ,the formula will become
f= 1/2L[tex]\sqrt{2T/meu}[/tex]
Hence the change in the fundamental frequency will be of the factor [tex]\sqrt{2}[/tex]
Therefore, the change will be of [tex]\sqrt{2}[/tex]
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suppose you want to observe the apparent retrograde motion of mars. how long will you need in order to see it?
Mars Retrograde Happens Every Two Years
As seen from the Earth(observer), the planets don't always make steady progress along the ecliptic(Path of the planets). They usually move in the same direction as the Sun (i.e., eastward), but from time to time they seem to slow down, stop, and reverse direction (i.e., move westward). This apparent reversal is called retrograde motion.
Apparent retrograde motion occurs when the observer (Earth) and the retrograding object (e.g., Mars) align with the Sun. Retrograde motion is an illusion created when we observe other planets from the moving planet Earth.
Example:
A familiar example of retrograde motion occurs when you pass a car on the freeway, the car being passed appears to move backwards relative to you.
For the planet mars, it takes almost two years to observe this kind of motion.
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a negative charge is placed at the center of a ring of uniform positive charge. what is the motion (if any) of the charge? what if the charge were placed at a point on the axis of the ring other than the center?
The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
if the charge were placed at a point on the axis of the ring other than the center the electric field then there is no motion.
electric field - It is a force produced by any charge near its surrounding.
If the charge were placed at a point on the axis of the ring other than the center then there is no motion because at the center of the ring E.F ( Electric field is zero.
electric filter the point on the axis of uniformly charged ring depend on the distance of the point from the center of the ring.
.
E.F = 0 because each half cancels each other . the negative charge will be in equilibrium as every part of the ring is uniformly attracted by it.
The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
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Compare Ampère's law with the Biota-Savart. law. Which is more generally useful for calculating →B for a current carrying conductor?
Both the law and its application make advantage of closed loops and symmetries.
Comparing Ampere's law to Biot-Savart law, the former is simpler to apply. In more complicated situations when symmetry cannot be applied, the Biot-Savart law is used.
Ampere's law is more helpful and simpler to apply if there is symmetry.
Gauss's law is used to compute the electric field, while Ampere's law is used to calculate the magnetic field.
What does the Biot-Savart law mean, and why is it important?The significance of the Biot-Savart law is as follows:
In electrostatics, Biot-Savart law is comparable to Coulomb's law.The law also applies to conductors that convey current but are very small in size.The law is applicable when the distribution of current is symmetrical.Both Ampere's law and the Biot-Savart law are applied to determine the magnetic field generated by a conductor that is carrying current. When there is symmetry present, we often employ Ampere's law.
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A reconnaissance plane flies 615 km away from
its base at 918 m/s, then flies back to its base
at 1377 m/s.
What is its average speed?
Answer in units of m/s.
The average speed of the plane is 3,967.7km/h or 1,102.15m/s.
Speed is a scalar number in kinematics that quantifies how quickly an item is moving. Since the plane's speed might fluctuate, we refer to the plane's average speed when describing its rate of motion. The whole distance travelled is divided by the total amount of time required to go that distance to arrive at this pace.
[tex]speed = \frac{distance}{time}[/tex]
For, [tex]time (t_{1} ) = \frac{615}{918}[/tex]
as 1m/s = 3.6km/h , therefore,
[tex]time (t_{1} ) = \frac{615}{918*3.6}[/tex]
[tex]time (t_{1} ) = 0.186 hours[/tex]
For, [tex]time(t_{2} ) = \frac{615}{1377}[/tex]
as 1m/s = 3.6km/h , therefore,
[tex]time (t_{2} ) = \frac{615}{1377*3.6}[/tex]
[tex]time (t_{2} ) = 0.124 hours[/tex]
Thus, the average speed of plane is,
[tex]s = \frac{615+615}{0.124+0.186}[/tex]
s = 3967.7km/h
or, [tex]\frac{3967.7}{3.6}[/tex] m/s = 1,102.15m/s
Therefore, the average speed of the plane is 3,967.7km/h or 1,102.15m/s.
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Calculate the centripetal force exerted on the earth by the sun. Assume that the period of revolution for the earth is 365. 25 days, the average distance is 1. 5 × 108 km and the earth’s mass is 6 × 1024 kg.
The centripetal force exerted on the earth by the sun is 3.56775 × 10²² N.
Mass of the Earth, m = 6 × 10²⁴ kg
The average distance between the Sun and the Earth is, r = 1.5 × 10⁸ km . r = 1.5 × 10¹¹ m
The path of the Earth around the Sun is circular, with a circumference:
c = 2πr
c = 2 × 3.1416 × 1.5 × 10¹¹ m
c = 9.42478 × 10¹¹ m
The Earth takes time of T = 365.25 days to complete one revolution around the Sun, that is to travel the distance of c.
T = (365.25 days) × (24hr/1day) × (60min/1hr) × (60s/1min)
T = 3.15576 × 10 ⁷s.
Thus the velocity of the Earth is,
v = c/T
v = (9.42478 × 10¹¹ m) / (3.15576 × 10 ⁷s)
v = 2.98653 × 10⁴ m/s
The centripetal force exerted on the Earth by the Sun is given by,
F = 2mv/r
F = 2× (6 × 10²⁴ kg) × (2.98653 × 10⁴ m/s ) /(1.5 × 10¹¹ m)
F = 3.56775 × 10²² N
Therefore, the centripetal force exerted on the earth by the sun is 3.56775 × 10²² N
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a wet bicycle tire leaves a trace of water on the floor.the tire has a radius of 30 cm,and the bicycle wheel makes 3 full rotations before stopping.
565.2 cm trace of water left on the floor
Given that,
A wet bicycle tire leaves a trace of water on the floor
The tire has a radius of 30 cm. and the bicycle wheel makes 3 full rotations before stopping
The trace of water left on the floor is the circumference of cycle
The circumference of circle is:
r = 30 cm
Therefore,
Given that,
bicycle wheel makes 3 full rotations before stopping
trace of water = 3 x 188.4 = 565.2
Thus, 565.2 cm trace of water left on the floor.
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An electric field of magnitude 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350m wide and 0.700m long (c) if the plane contains the y axis and its normal makes an angle of 40.0° with the x axis.
The value of electric flux for mentioned plane will be 657N⋅m²/C at 40° with the x axis.
For a uniform electric field passing through a plane surface,
electric flux(ϕ)=EAcosθ
where,
θ is the angle between the electric field and the normal to the surface,
E is the electric field
A is surface area
(a) The electric field is perpendicular to the surface so θ=0
=(3,50×10³ N/C)[(0.350m)(0.700m)]cos0
=858N⋅m²/C
(b)
The electric field is parallel to the surface θ=90
, so cosθ=0
therefore,
the flux is zero.
(c) For the mentioned plane,
electric flux(ϕ)=(3.50×10³N/C)[(0.350m)(0.700m)]cos40.0
=657N⋅m²/C
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a block weighing 10 lbf and having dimensions 10 in. on each edge is pulled up an inclined surface on which there is a film of sae 10w oil at 100c f. if the speed of the block is 2 ft/s and the oil film is 0.001 in. thick, find the force required to pull the block. assume the velocity distribution in the oil film is linear. the surface is inclined at an angle of 25° from the horizontal.
The force required to pull the block will be 13.89 kN.
What is the force?Let μ be the viscosity of oil, y be thickness, v be the velocity of the block, and A be the base area of the block. Then the force is given as,
F = μAv / y
An inclined surface with a coating of sae 10w oil at 100°C is dragged up by a block that weighs 10 lbf and measures 10 in on each edge. if the oil coating is 0.001 in. thick and the block is moving at a speed of 2 ft/s. Assume that the oil film's velocity distribution is linear. A 25° angle from the horizontal separates the surface from the horizontal.
The diagram is given below. Then the force is given as,
F - 10 sin 25° = 10 × (10 / 12)² × 2 / 0.001
F = 4.226 + 13888.89
F = 13893.1 N
F = 13.89 kN
13.89 kN of force will be required for pulling the block.
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Suppose you are transported to a planet with twice the mass of earth, but the same radius of earth. Your weight would __________ by a factor of __________.
Answer:تاتا البال،ًـلاللبببلا
لولاليثقغاايصسضشطذدزلب
Explanation:ابرلرتهىغرلب
Consider a solenoid that is very long compared with its radius. Of the following choices, what is the most effective way to increase the magnetic field in the interior of the solenoid? (a) double its length, keeping the number of turns per unit length constant (b) reduce its radius by half, keeping the number of turns per unit length constant (c) overwrap the entire solenoid with an additional layer of current-carrying wire
Increasing the coil's number of turns expanding the current. Putting a core made of iron into the solenoid.
(a) Because Bpropto n states that the magnetic field will grow if the number of turns per unit length is increased.
(b) Once more, B=mu 0(H+M)=mu 0(nI+M) is true, with M being the magnetization of the magnetic material. Therefore, the magnetic field will intensify if we place a magnetic substance within the solenoid.. (correct option)
c) because, according to Bpropto I, the magnetic field will increase if we raise the current.. (correct option)
(d) Because Bpropto frac1 L, if we lengthen it but keeping the total number of turns same , the magnetic field will decrease....(option (d) is incorrect)
(e) If we keep the number of turns per unit length constant , the magnetic field will unchanged ....(option (e) is incorrect )
what is referred by the cross sectional area of a solenoid?The way it is wound, or more specifically, the circle, is referred to as the cross section. Circle does not depend on what it has within.
A solenoid is a wire wound in general in a circular shape, even when it is not forced, the cross section refers to the way in which it is wound, in general to the circle and is joined to determine the area A = π r². This circle is the same whether it is in the air, a liquid, or a solid because it is independent of what is inside,
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two small charged objects repel each other with a force 2.53 when separated by a distance 0.11. if the charge on each object is reduced to 67.9% of its original value and the distance between them is reduced to 0.053 the force becomes
The force between the charges will be 10.9 N.
What is the electrostatic force?The electrostatic attraction particles that make up the electrostatic force behave as both an attracting and a repulsive force. The force is inversely proportional to the square of the distance between them.
F ∝ 1 / r²
F = k / r²
Fr² = k
When two tiny charged objects are 0.11 of a distance apart, they repel one another with a force of 2.53. If the distance between the two particles is cut to 0.053 and the charge on each item is decreased to 67.9% of its initial amount. Then the force will be
F₁r₁² = F₂r₂²
2.53 × (0.11)² = F × (0.053)²
F = 10.9 N
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the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.
Three-dimensional shape the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.
What do you mean by three-dimensional shape?A solid figure, object, or shape with three dimensions—length, breadth, and height—is known as a three-dimensional shape in geometry. Three-dimensional shapes have height, which is equivalent to thickness or depth, unlike two-dimensional shapes. These objects are sometimes referred to as 3-D forms because three dimensions can also be written as 3-D. Every 3-D shape takes up space, which is quantified by volume. For instance, The fundamental 3-dimensional shapes we encounter every day include a cube, rectangular prism, sphere, cone, and cylinder. We can see 3-D shapes all around us. A square prism can be seen in a book and a box, a cube in a Rubik's Cube and a die, etc.
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what is the magnitude (in n/c) and direction of an electric field that exerts a 2 x 10-5 n upward force on a -1.2 µc charge? 1µc
|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.
We have a Charge particle present inside a Electric field.
We have to determine the magnitude (in n/c) and direction of an electric field that exerts a 2 x [tex]10^{-5}[/tex] Newton.
What is Electric Field Intensity?The electric field intensity at a point R meters away from a charge Q is the amount of force experienced by the test charge ([tex]$q_{o}[/tex]) of unit magnitude at a distance R meters away the Charge Q. Mathematically -
E = [tex]$\frac{F}{q}[/tex]
According to the question, we have -
F = 2 x [tex]10^{-5}[/tex] Newton
|Q| = - 1.2 [tex]\mu[/tex]C = - 1.2 x [tex]10^{-6}[/tex] [tex]\mu[/tex]C = 1.2 x [tex]10^{-6}[/tex] [tex]\mu[/tex]C
Therefore -
E = [tex]$\frac{2\times 10^{-5} }{1.2\times 10^{-6} }= \frac{100}{6} = 16.67\;N/C[/tex]
Therefore -
|E| = 16.67 N/C and the direction of this Electric field is in the downward direction.
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Explain grounding in your own words
Explain why two parallel wires carrying currents in opposite directions repel each other.
As in a series circuit, if you have two parallel wires with current flowing in opposite directions, the magnetic fields of the two wires will be moving in the same direction at the point of intersection and will repel one another.
What function does parallel wiring provide in an electrical circuit?In a parallel circuit, the branches divide the current so that only a portion of it passes through each branch. In a parallel circuit, each branch has the same voltage or potential difference, but the currents may differ. For instance, in a residential electrical circuit, the same voltage is provided across each lamp and appliance, but depending on their power requirements, each load draws a varied amount of current. A group of comparable batteries connected in parallel can produce more current than a single battery while maintaining the same voltage.
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Study the graph below, it represents a motion with 3 distinct parts. Pay particular attention to label on the y-axis. Using at least 1 complete sentence, describe what is happening during each part of the motion. [your answer must include words and phrases that describe motion. Suggested examples are things like accelerating, constant velocity, backward, forward, at rest, stop]
We describe the motion in the graph as
In the first 2 seconds, the motion of the object is that it at rest. From t = 2 seconds to t = 4 seconds, the motion of the object is that it accelerating. For t = 4 seconds to t = 6 seconds, the motion of the object is that it is moving with constant velocityWhat is motion?Motion is the change in position of an object with time.
How to describe the motion in the graph?Since on the graph
the y - axis is represented by velocity and the x - axis is represented by time,Motion from t = 0 to t = 2 secondsFrom the graph, in the first 2 seconds, the motion of the object is described as the object is at rest.
Motion from t = 2 to t = 4 secondsFrom the time , t = 2 seconds to t = 4 seconds, we see that the graph has a positive slope. During this time interval, from t = 2 seconds to t = 4 seconds, the object is accelerating.
So, the motion of the object is described as the object is accelerating
Motion from t = 4 to t = 6 secondsFor the time interval t = 4 seconds to t = 6 seconds, we see that the velocity is a straight line parallel to the x(time) axis. This implies that the velocity of the object is constant for t = 4 seconds to t = 6 seconds. So, for t = 4 seconds to t = 6 seconds, the motion of the object is described as object is moving with constant velocity
So, we describe the motion in the graph as
In the first 2 seconds, the motion of the object is that it is at rest. From t = 2 seconds to t = 4 seconds, the motion of the object is that it is accelerating. For t = 4 seconds to t = 6 seconds, the motion of the object is that it is moving with constant velocityLearn more about motion graph here:
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two charges experience exert a force of 1n on each other when they are 1m apart. what force will these charges experience if they are placed 2m apart?
The force experienced by the charges when they are placed 2 m apart will be 0.25 N.
The force between two charges is given by Coulomb’s Law.
According to this law, the force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges.
The above statement is represented by the following equation,
F= K Q1 x Q2/r^2
Here, F = Force between the two charges
K = Constant
Q1 and Q2 = Magnitude of the two charges
r = Separation between the two charges
According to this equation, the force is inversely proportional to the square of the distance between two charges.
Initially, when the two charges are 1 m apart, the force is 1 N.
When the distance between the two charges is doubled,
then according to Coulomb’s Law, the force should decrease by 4 times.
Hence, if the charges are placed 2 meters apart, the force becomes 1/4= 0.25 N.
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A driver travels northbound on a highway at a speed of 25.0v m/s. A police car, traveling southbound at a speed of 40.0 m/s , approaches with its siren producing sound at a frequency of 2500 Hz. (a) What frequency does the driver observe as the police car approaches?
The driver observes as the police car approaches [tex]f^{\prime}=3036 \mathrm{~Hz}[/tex]
An observer traveling at speed is a driver [tex]v_O=25.0 \frac{\mathrm{m}}{\mathrm{s}}[/tex] and the source is the swiftly moving police vehicle [tex]v_S=40.0 \frac{\mathrm{m}}{\mathrm{s}}[/tex].The frequency of the sound that the source emits is [tex]f=2500Hz[/tex] The speed of the sound is [tex]v=343 \frac{\mathrm{m}}{\mathrm{s}}[/tex].
(a) The source and the observer approach one another. Then, we apply [tex]13.30[/tex]
[tex]f^{\prime}=f \frac{v+v_O}{v-v_S}[/tex]substituting numerical numbers for the positive values of [tex]V_O[/tex] and [tex]V_S[/tex]
[tex]\begin{gathered}f^{\prime}=f=2500 \mathrm{~Hz} \times \frac{343 \frac{\mathrm{m}}{\mathrm{s}}+25.0 \frac{\mathrm{m}}{\mathrm{s}}}{343 \frac{\mathrm{m}}{\mathrm{s}}-40.0 \frac{\mathrm{m}}{\mathrm{s}}} \\f^{\prime}=3036 \mathrm{~Hz}\end{gathered}[/tex]
Hence, the driver observes as the police car approaches [tex]3036 Hz[/tex] is the frequency.
Explain reason for this observation?When a wave source is moving relative to an observer, the Doppler effect or Doppler shift is a phenomenon that can be seen. A typical physical illustration of the Doppler Effect is when an ambulance crosses you while its siren is on. Let's take a closer look at the Doppler effect's nuances.
The Doppler effect is a significant phenomenon in planetary science and other scientific fields. The frequency of any sound or light wave produced by a moving source with regard to an observer is said to fluctuate as a result of the Doppler effect or Doppler shift.
A source's waves are compressed as they move in the direction of the observer. In contrast, waves from a source that is moving away from the observer become longer.
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Pleaseeeeeee help me out.
The give scenarios illustrates the following laws of motion respectively:
Pushing a baby on a swing is easier than pushing an adult on the same swing - Second law Water spills from a glass carried by someone who is walking steadily and suddenly stops short - First lawWhen thrown with the same force, a soccer ball accelerates more than a bowling ball - Second lawA magician pulls a tablecloth out from under a dish on a table without disturbing the dish - Third lawA rocket launches into space, pushing fuel exhaust in one direction and the rocket in the opposite direction - Third lawA book rests on top of a shelf and does not move until a student accidentally knocks it off - First lawWhat are the three laws of motion?The three laws of motion are stated below as follows:
First law - it states that a body continues in its state of rest or uniform motion in a straight line unless acted upon by an external force. This is also known as the law of inertia.
Second law - it states that the rate of change of the momentum of a body is directly proportional to the applied force and takes place in the direction of the applied force. This law is restated as acceleration is proportional to force and inversely proportional to mass.
Third law - for every force, there is an equal and oppositely directed reaction.
Considering the individual scenarios:
Pushing a baby on a swing is easier than pushing an adult on the same swing - Second law
Water spills from a glass carried by someone who is walking steadily and suddenly stops short - First law
When thrown with the same force, a soccer ball accelerates more than a bowling ball - Second law
A magician pulls a tablecloth out from under a dish on a table without disturbing the dish - Third law
A rocket launches into space, pushing fuel exhaust in one direction and the rocket in the opposite direction - Third law
A book rests on top of a shelf and does not move until a student accidentally knocks it off - First law
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If the mass of a planet is 3. 10 1024 kg, and its radius is 2. 00 106 m, what is the magnitude of the gravitational field, g, on the planet's surface?
The gravitational field strength is 51.6925 N/kg.
We need to know about the gravitational field to solve this problem. The gravitational field is the area where affected by gravitational force. The magnitude of the gravitational field can be calculated by this equation
g = G . m / r²
where g is the gravitational field, G is gravitational constant (6.67 x 10¯¹¹ Nm²/kg²), m is the mass of the planet and r is the radius of the planet
From the question above, we know that
m = 3.10 x 10²⁴ kg
r = 2.00 x 10⁶ m
By substituting the given parameter, we get
g = G . m / r²
g = 6.67 x 10¯¹¹ . 3.10 x 10²⁴/ (2.00 x 10⁶)²
g = 51.6925 N/kg
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f a ball is thrown into the air with an initial velocity of 50 ft/s, its height in feet after t seconds is given by y
The instantaneous velocity of the ball is 18ft/s.
What is velocity?
The pace at which an object's position changes as perceived from a particular point of view and as measured by a particular unit of time (for example, 60 km/h northbound) is defined as its velocity. Velocity is a fundamental concept in kinematics, the branch of classical mechanics that deals with the motion of bodies.
In order to be defined, the physical vector quantity known as velocity needs to have both a magnitude and a direction. Speed is a coherent derived unit whose quantity is measured in metres per second (m/s or m/s1) in the SI. Speed is the scalar absolute value (magnitude) of velocity (metric system).
Calculations:
at t=1s, the height is y=50(1)-16(12)=34ft.
at t=1.1s, the height is y=50(1.1)-16(1.12)=35.64ft
Between these two heights, the change in time is 0.1s. The change in height is 35.64ft-34ft=1.64ft.
Therefore [tex]V_{avg}=\frac{del y}{del t} = \frac{1.64ft}{0.1s} =16.4ft/s[/tex]
at t=1.01s, the height is y=50(1.01)-16(1.012)=34.1784ft
between t=1.01s and t=1s the change in time is 0.01s and the change in height is 0.1784ft.[tex]V_{avg}=\frac{0.1784ft}{0.01s} =17.84ft/s[/tex]
at t=1.001, the height is y=50(1.001)-16(1.0012)=34.017984ft
between t=1s and t=1.001s, the change in time is 0.001s and the change in height is 0.017984ft
[tex]V_{avg} =\frac{0.017984ft}{0.001s}=17.984ft/s[/tex]
It appears that as the time interval approaches 0, the average velocity approaches 18ft/s. Therefore it is safe to assume the instantaneous velocity at t=1s is 18ft/s.
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A meteor is falling to earth and enters the Earth’s atmosphere at 0=−435 ŷ, if it experiences an acceleration due to Earth’s gravity of =−9.8 2ŷ, how what will its velocity be after t= 13s? (assume all motion is vertical, which means on the y-axis. You will use ŷ for up, and - ŷ for down)
At 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s
We know that,
v = u + at
where,
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
Given that,
u = - 435 m / s
a = - 9.8 m / s²
t = 13 s
v = - 435 - ( 9.8 * 13 )
v = - 562.4 m / s
Velocity is the rate of change of distance with respect to time. It is denoted by V.
Therefore, at 13 seconds, the meteor's velocity is - 562.4 m / s, if it enters the earth's atmosphere at - 435 m / s
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determine the car’s instantaneous velocity at t=5s
Answer:
In the given graph,
Velocity is constant between 4 to 8 which is 2m/s
So, at t=5s
Instantaneous velocity=2m/s
Explanation:
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