A student walks 15 meters west, then turns south and walks for 7 meters. The student then begins to walk east for 15 meters until he reaches his destination. What was the student's distance? What about displacement? (hint draw it)

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.

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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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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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.

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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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.

The force the earth exerts on you is calculated as follows;

F = mg

where;

F = 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.

F = GmM/R²

where;

F = (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

F = 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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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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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.

The significance of the Biot-Savart law is as follows:

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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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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The force between the charges will be 10.9 N.

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 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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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:

hope this help u thank u

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]

Disclaimer:  the question has missing part 1 and part 2 and hence part 1 has not been answered

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|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.

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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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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The give scenarios illustrates the following laws of motion respectively:

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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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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 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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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 )

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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Three-dimensional shape the presence of suggests that an object has weight and gravity, while the absence of it suggests lightness or airiness.

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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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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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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Answer:تاتا البال،ًـلاللبببلا

لولاليثقغاايصسضشطذدزلب

Explanation:ابرلرتهىغرلب

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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The force required to pull the block will be 13.89 kN.

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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A controlled burn removes dead vegetation that might otherwise help a wildfire start and spread.

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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The instantaneous velocity of the ball is 18ft/s.

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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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]

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.

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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We describe the motion in the graph as

Motion is the change in position of an object with time.

Since on the graph

From the graph, in the first 2 seconds, the motion of the object is described as the object is at rest.

From 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

For 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

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The student's shoulder supports the weight of the bag.

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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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

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

Vertical velocity = u × Cosθ

Vertical velocity = 2×10³ × Sin30

Vertical velocity = 1×10³ m/s

3. How to determin the range

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

H = u²Sine²θ / 2g

H = [(2×10³)² × (Sine 30)²] / (2 × 9.8)

H = 1.53×10⁵ m

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