The gravitational force between two objects is 45,000 N. What will be the gravitational force between the two objects if the distance between the objects is reduced by one-half?

The direction of the magnetic field at a point in an electromagnetic wave can be determined using the 'right-hand rule'. Point your thumb in the direction of the electric field and curl your fingers. The direction your fingers curl is the direction of the magnetic field.

In physics, particularly electromagnetic theory, the electric and magnetic fields of an electromagnetic wave are always perpendicular to each other and to the direction of propagation. The direction of the magnetic field at point 'a' can be determined by using the 'right-hand rule'. This rule states that if you point your thumb in the direction of the electric field and curl your fingers, the direction your fingers curl is the direction of the magnetic field. So, the direction of the magnetic field at point 'a' would be in the direction your fingers curl when you point your thumb in the direction of the electric field at point 'a'.

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Answer:

the answer is A (×-5)(×-1)

Answer:

A. (x - 5)(x - 1)

Explanation:

x² - 6x + 5

ax² + by + c

What 2 numbers give a product of 5 (a * c) and a sum of -6 (b)?

-5, -1 ---> -5 + -1 = -6, (-5) * (-1) = 5

x² - 6x + 5 = (x - 5)(x - 1)

Does that help?

The upward force exerted on an object falling through air is air resistance. The correct option is C.

When an object moves through air, it creates a force called air resistance. This force basically acts in the inverse way of direction of a body moving via air.

The frictional force of air resistance acts on the moving body. When a body moves, air resistance slows it down.

When an object moves through the air, it encounters air resistance. Resistance varies depending on the velocity, shape, and area of the object. The higher the air resistance, the faster an object moves and the larger its area.

When air pushes against a moving object, it creates air resistance force. Frictional forces include air resistance. Force is always applied against the motion of an object.

Thus, the correct option is C.

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The magnetic field produced by a current loop is not uniform. It is strongest at the center of the loop and decreases as you move away from the center. This change in strength across the space surrounding the loop makes it non-uniform.

No, the magnetic field created by a current loop is not uniform. A current loop means a closed electrical circuit or a current flowing in a loop of conducting wire. When a current flows through the loop, a magnetic field is generated surrounding the loop. This magnetic field is strongest at the center of the loop and diminishes as you move away from the center. This variance in strength implies that the magnetic field is not uniform.

Example:

If you place a compass around a loop with flowing current, you'd notice that the compass needle's direction (which aligns with the magnetic field) changes as you move the compass around. This indicates that the magnetic field strength (and direction) changes across the space surrounding the current loop, which makes it non-uniform.

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Answer:

0.00156875 m

Explanation:

y = Distance from the 20th fringe = 9.6 mm

l = Distance to screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = [tex]502\ nm[/tex]

m = Number of fringe = 20

We have the relation

[tex]\dfrac{y}{l}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda l}{y}\\\Rightarrow d=\dfrac{20\times 502\times 10^{-9}\times 1.5}{9.6\times 10^{-3}}\\\Rightarrow d=0.00156875\ m[/tex]

The seperation of the two slits is 0.00156875 m

Answer: True

Explanation:

A photo detector that can respond to the entire rang of visible light can be design, it is true.

Photo detector is a device in an optical receiver which receives optical signals and convert it to electric signal. It is the key device position in front of the optical receiver.

Explanation:

The direction of magnetic field in a straight wire is given by Maxwell's right hand rule. It states that if the thumb of right hand points towards the direction of electric current, then the curled finger will give the direction of magnetic field.

In this case, a vertical wire carries a current straight down. It means that current will flow in the clockwise direction.

The magnetic field to the east of a vertical wire carrying a current straight down is directed into the page, which is determined by the right-hand rule.

To determine the direction of the magnetic field around a current-carrying wire, we can use the right-hand rule. For a wire that carries a current downward, you would point the thumb of your right hand in the direction of the current (downward), and then your fingers would naturally curl in the direction of the magnetic field. To the east of the wire, your fingers would point into the page, indicating that the magnetic field is directed into the page.

Final answer:

In the provided scenario, the correct description of the body in terms of simple machines is that the lower legs are levers and the knees serve as fulcrums.

Explanation:

When we analyze the biomechanics of the body using the piece of gym equipment in the provided scenario, particularly during a leg workout, we can compare the body to a set of simple machines. In this context, the lower legs act as levers.

The knees operate as fulcrums, which are the points about which the levers rotate. The ankles do not carry or hold the load in this action; rather they act as the point of application for the effort when extending the foot in movements like calf raises.

The configuration described is analogous to a first-class lever, where the fulcrum (knee) is situated between the effort (muscles applying force near the ankle) and the load (the resistance on the leg extension machine). Hence, the correct description of the body in terms of simple machines is: "The lower legs are levers, and the knees are fulcrums."

To address the components from the question:

The knees are the fulcrums since they are the pivot point for the motion of the lower leg.

The lower legs are levers because they move around the fulcrum to lift the load on the gym equipment.

The ankles are where the calf muscles apply the effort but are not where the load is held.

Therefore, third option is correct: The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Answer:

[tex]\lambda=4L=1.33m[/tex]

v=343m/s

Explanation:

We have to take into account the expressions

[tex]f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}[/tex]

if we assume that 256Hz is the fundamental frequency we have

[tex]f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m[/tex]

and for wavelength

[tex]\lambda=4L=1.33m[/tex]

hope this helps!!

Answer:

a) The wavelength, λ = 1.352 m

b) The speed of the sound in the column air is v = 346.112 m/s

Explanation:

a) Frequency, f = 256 Hz

The water level is lowered by 676 mm. i.e. x = 676 * 10⁻³ m

The length, x, at which the loudest sound will be heard is λ/2

i.e. x =  λ/2

Therefore the wavelength of the 256 Hz sound waves in the column air is given by the relation

λ = 2x

λ = 2 * 676 * 10⁻³

λ = 1.352 m

b) The speed of the sound in the column air:

The speed, v, is given by the relation   v = f  λ

speed, v = 256 * 1.352

v = 346.112 m/s

(C) Air Resistance

Explanation:

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

Final answer:

Sound energy is a form of kinetic energy, not potential energy, because it arises from the motion and vibration of particles within a medium as the sound wave propagates.

Explanation:

Sound energy is not considered a type of potential energy because it is inherently associated with the motion of particles; therefore, it is a form of kinetic energy. Sound is a mechanical wave that results from the vibration of particles within a medium. As these particles oscillate, they transfer energy through the medium, resulting in what we perceive as sound. While a sound wave indeed combines kinetic and potential energy due to the motion of particles and the elasticity of the medium, the propagation of the wave itself and the sound energy we detect are manifestations of kinetic energy. In the case of potential energy, this form of energy is related to an object's position or state; for instance, a rock atop a hill has gravitational potential energy due to its elevated position in a gravitational field. Similarly, the chemical energy in a battery is potential energy based on the chemical composition and potential for a chemical reaction.

To help visualize the concept, consider water behind a dam. The water has potential energy because of its elevated position. As it moves and flows through turbines, it converts this potential energy into kinetic energy, which then can be harnessed to produce electricity. In contrast, when a guitar string is plucked, the energy initially stored in the deformed string is potential energy, but as the string returns to its normal state and vibrates, it generates sound—which is an expression of kinetic energy. Thus, when sound is produced, it is the motion and transfer of energy through a medium, reflecting kinetic, not potential, energy characteristics.

Answer:

they have a cooling effect! :)

Explanation:

Answer:

A cooling effect

Explanation:

As strange as it seems, volcanoes cool during eruption. All of the magma pent up in the core gets released into lava, which drastically changes the internal temperatures. This is kind of like how for some people, yelling/beating on inanimate objects helps them to "cool off".

Answer:

Explanation:

Hi

The net force on a person in a descending elevator can be calculated using Newton's second law of motion, considering the person's weight and the elevator's downward acceleration.

The problem relates to Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In the context of a person in a descending elevator, the net force exerted on the person by the floor, sometimes referred to as the normal force, can be found by considering both the gravitational force (weight) acting downwards and the acceleration of the elevator.

Given the person's weight (W) is 686 N and the downward acceleration (a) of the elevator is 1.8 m/s², we can first find the person's mass (m) by rearranging the formula for weight (W = mg), where g is the acceleration due to gravity (9.8 m/s²), giving us m = W/g. Secondly, we can calculate the net force (Fnet) using Newton's second law, taking into account that the actual acceleration of the person is the difference between the acceleration due to gravity and the elevator's acceleration since the elevator is moving downward. Therefore, Fnet = m(g - a).

Net force acting on a person can be calculated as the difference between the force of gravity acting on the person and the force exerted by the elevator. First, calculate the force of gravity: Fg = mg, where m is the mass and g is acceleration due to gravity (9.8 m/s²). Then, calculate the force exerted by the elevator using the equation Fnet = ma, where a is the acceleration of the elevator. Finally, find the net force by subtracting the force of gravity from the force exerted by the elevator.

Answer:

Explanation:

We have that for the Question "Imagine two circular plates; one is solid and the other has a hole cut out of the center. Both plates have the same radius, same thickness, and same mass. The same force F is applied tangential to the edge of each plate in such a way that the plates rotate about an axis passing through the center and perpendicular to the surface of the plates."

Question: Which one of the following statements is true regarding the angular acceleration?

Answer : The solid plate will have the greater angular acceleration

Question : Which of the following statements helps to explain the question asked above? (Select all that apply.)

Answer :

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Answer:

The time that will have to pass before one has under 1 oz of caffeine remaining is 24.53 hours

Explanation:

Here, we have the formula for half life as follows;

[tex]N(t) = N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}[/tex]

Where:

N(t) = Remaining quantity of the substance = 1 oz

N₀ = Initial quantity of the substance = 30 oz

t = Time duration

[tex]t_{1/2}[/tex] = Half life of the substance = 5 hours

Therefore, plugging in the values, we have

[tex]1= 30(\frac{1}{2})^{\frac{t}{5}}[/tex]

[tex]\frac{1}{30} = (\frac{1}{2})^{\frac{t}{5}}\\ln(\frac{1}{30}) =\frac{t}{5} ln(\frac{1}{2})\\\frac{t}{5} = \frac{ln(\frac{1}{30}) }{ ln(\frac{1}{2})} = 4.91\\ t = 4.91 \times 5 = 24.53 \ hours[/tex]

The time that will have to pass before one has under 1 oz of caffeine remaining = 24.53 hours.

Answer:

[tex]F = 22298.824\,N[/tex]

Explanation:

According to the Principle of Energy Conservation and the Work-Energy Theorem, the bullet has the following expression:

[tex]U_{g,A} + K_{A} = U_{g,B} + K_{B} + W_{loss}[/tex]

[tex]W_{loss} = U_{g,A}-U_{g,B} + K_{A}-K_{B}[/tex]

[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot [v_{A}^{2}-v_{B}^{2}][/tex]

The average force exerted on the bullet to stop it is:

[tex]F = \frac{m\cdot [v_{A}^{2}-v_{B}^{2}]}{2\cdot \Delta s}[/tex]

[tex]F = \frac{(7.8\times 10^{-3}\,kg)\cdot [(540\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}]}{2\cdot (0.051\,m)}[/tex]

[tex]F = 22298.824\,N[/tex]

Answer:

22298.82N

Explanation:

The bullet has a kinetic energy: ½*m*v²

mass of bullet = 7.80g = 7.80÷1000 = 0.0078kg

distance in meter =5.10cm = 5.10÷100 = 0.051meter

K.e = 0.5 ×0.0078 × 540× 540

K.e = 1137.24 J

When the bullet stops this energy goes to zero, as the energy must be conserved the work done by the head of the superhero must be equal to the original energy of the bullet:

W= K

Considering an average constant force, the work can be calculated as:

W=F*d=K

Solving for F:

F = K/d

1137.24 J/ 0.051m

= 22298.82N

The magnitude of the normal force and acceleration is calculated using Newton's second law and free body diagrams, considering all forces acting on the object.

The question pertains to finding the magnitude of the acceleration and the normal force acting on a chair, block, or similar object using Newton's second law of motion. To determine these quantities, one must analyze the forces acting on the object in a free body diagram and apply the equations of motion. The normal force is the force exerted by a surface to support the weight of an object resting on it, and it acts perpendicular to the surface. To find the normal force (FN), we typically assume static equilibrium if the object is at rest, which implies that the net force in the vertical direction is zero; thus, FN equals the object's weight (mg), where m is the mass and g is the acceleration due to gravity. The magnitude of acceleration (a) can be calculated using the net force acting on the object (resultant force) divided by its mass, as stated by Newton's second law (F = ma).

For example, if a block is placed on a table, the normal force is equal to the gravitational force acting on it (FN = Fg). If the block is accelerating, you would calculate the net force by including all forces acting on the block, such as tension, friction, or applied forces, and then use F = ma to find the acceleration.

Answer:

I₂ = I₀  1/4

Explanation:

For this exercise we use the definition of intensity which is the power per unit area

         I = P / A

The emitted power is constant, so

     P = I A

We can write this equation for the start and end point with index 2

    I₀ A₀ = I₂ A₂

    I₂ = I₀ A₀ / A₂

The spot area is the area of ​​the circle

      A₀ = π r₀²

We substitute

     I₂ = I₀ r₀² / r₂²

It indicates that the radius of the spot is twice the initial radius

      r₂ = 2 r₀

      I₂ = I₀ (r₀ / 2 r₀)²

      I₂ = I₀  1/4

a) 0.0028 rad/s

b) [tex]23.68 m/s^2[/tex]

c) [tex]0 m/s^2[/tex]

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

[tex]\omega = \frac{\theta}{t}[/tex]

where

[tex]\theta[/tex] is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

[tex]\omega = \frac{v}{r}[/tex] (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

[tex]v=29,960 km/h[/tex] is the linear speed, converted into m/s,

[tex]v=8322 m/s[/tex]

[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

[tex]\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s[/tex]

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

[tex]a_r=\omega^2 r[/tex]

where

[tex]\omega[/tex] is the angular speed

[tex]r[/tex] is the radius of the orbit

For the spaceship in the problem, we have

[tex]\omega=0.0028 rad/s[/tex] is the angular speed

[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

[tex]a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2[/tex]

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

[tex]a_t=\frac{\Delta v}{\Delta t}[/tex]

where

[tex]\Delta v[/tex] is the change in the linear speed

[tex]\Delta t[/tex] is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

[tex]v=8322 m/s[/tex]

Therefore, its linear speed is not changing, so the change in linear speed is zero:

[tex]\Delta v=0[/tex]

And therefore, the tangential acceleration is zero as well:

[tex]a_t=\frac{0}{\Delta t}=0 m/s^2[/tex]

Explanation:

this is because if your going to touch something dangerouse your hand normally feels it first and its the front not the back so when you pick things up you can tell if its dangerouse hope that helped

The palm of the hand has more touch receptors than the back, due to its glabrous skin type which is thicker and more sensitive. These receptors allow for detection of fine touch and pressure, critical for manual tasks.

Considering the density of touch receptors in different parts of our body, you could expect to have more touch receptors on the palm of your hand than on the back of your hand. This is due to the specific type of skin found on the palm, referred to as glabrous skin. This type of skin is typically thicker and more sensitive as compared to hairy skin, hence it contains more touch receptors.

The importance of the high density of touch receptors in areas such as palms and fingertips can be seen in the daily actions, such as the ability to sense fine touch and detailed information, which is necessary for fine motor tasks.

Furthermore, mechanoreceptors such as Merkel's disks, Meissner's corpuscles, Pacinian corpuscles, and Ruffini endings play a vital role in interpreting the touch-related sensory information of these regions. For instance, Merkel's disks and Meissner's corpuscles that are found deeper in the skin and in the upper parts of the skin respectively, detect fine touch. On the other hand, Pacinian corpuscles and Ruffini endings can sense deeper touch like pressure.

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Answer:

tipping over dominoes

Explanation:

?

One game that represents the concept of a wave is the puzzle game "Wave Wave."

In "Wave Wave," players control a small triangle as it navigates through a series of challenging and rhythmic levels filled with obstacles. The gameplay involves moving the triangle in sync with the pulsating waves that define the level's layout.

The game's visuals and mechanics are designed to mimic the appearance and behavior of a wave, with the obstacles and patterns resembling the peaks and troughs of waves. The player's goal is to maintain control and timing, moving the triangle smoothly through the undulating paths while avoiding collisions.

"Wave Wave" is a creative interpretation of the wave concept in a gaming context, with its gameplay mechanics directly tied to the idea of navigating through waves. Keep in mind that there might be other games that also incorporate wave-like mechanics, so this is just one example.

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Answer:

The final temperature is 16°C

Explanation:

mcΔT (water) = mcΔT (zinc)

0.50(4184)(t-15) = 0.040(388)(115-t)

2092(t-15) = 15.52(115-t)

2092t-31380 = 1784.8-15.52t

2029t+15.52t= 1784.8+31380

2044.52t = 33164.8

t = 33164.8/2044.52

t = 16

Answer: B. Both involve the use of levers.

Explanation:

Answer:

Force will be needed to make the object move

Explanation:

An object at rest will stay at rest and an object in motion will stay in motion unless acted on by an unbalaced object.

Answer:

magnetic field is the correct answer.

Explanation:

Answer:

Acceleration, a = [tex]2.8\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the skier, u = 8 m/s

Final speed of the skier, v = 6 m/s

Distance, d = 5 m

We need to find the magnitude of her acceleration. It is equal to the rate of change of speed. It is given by :

[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{6^2-8^2}{2\times 5}\\\\a=-2.8\ m/s^2[/tex]

So, the magnitude of her acceleration is [tex]2.8\ m/s^2[/tex].

The magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.

To find the magnitude of the skier's acceleration, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the initial velocity is 8.0 m/s, the final velocity is 6.0 m/s, and the time is 5.0 seconds.

Plugging these values into the formula, we get:

Acceleration = (6.0 m/s - 8.0 m/s) / 5.0 s = -0.4 m/s^2

So the magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.

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Answer:

t² ∝ r³

Explanation:

Let m be the mass of the sun and m₁ be the mass of the planet and r its distance from the sun. Its gravitational force of attraction equals its centripetal force. So,

Gmm₁/r² = m₁rω² = m₁r(2π/t)²  were t is the period of orbit of the planet

Gm/r² = 4π²r/t²

rearranging we ave

t² = (4π²/Gm)r³   since k = 4π²/Gm = constant

t² = kr³ and

t² ∝ r³

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

Answer:

Explanation:

Given that,

The peak wavelength is

λp = 475nm

Then, we want to find the temperature,

From Wein's displacement law,

When the maximum is evaluated from the Planck radiation formula, the product of the peak wavelength and the temperature is found to be a constant (k = 2.898 ×10^-3 mK)

So, applying this we have

λp•T = 2.898 ×10^-3

T = 2.898 × 10^-3 / λp

T = 2.898 × 10^-3 / 475 × 10^-9

T = 6101.05 K

The surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

Given the data in the question;'

Peak wavelength; [tex]\lambda _p = 475nm = 4.75 *10^{-7}m[/tex]

To determine the surface temperature the star, we use the expression for peak wavelength by Wien's Displacement Law:

[tex]\lambda _p = \frac{0.2898 * 10^{-2}m.K}{T}[/tex]

Where T is the surface temperature

We substitute our value into the formula and solve for for "T"

[tex]4.75*10^{-7}m = \frac{0.2898 * 10^{-2}m.K}{T}\\\\T = \frac{0.2898 * 10^{-2}m.K}{4.75*10^{-7}m}\\\\T = 6. 1 * 10^3K[/tex]

Therefore, the surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

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