Based on the assumption that a liquid conducting core and rapid rotation are both required for a magnetic field to operate, which terrestrial planets would you expect to have magnetic fields?

Answers

Answer 1

answer is earth. Earth is having everything that is required for a magnetic field to operate.

Answer 2

Based on the assumption that a liquid conducting core and rapid rotation are both required for a magnetic field to operate, only Earth have magnetic fields.

What are terrestrial planets?

Because of their compact, rocky surfaces akin to Earth's terra firma, the planets Mercury, Venus, Earth, and Mars are referred to as terrestrial. The four planets closest to the sun are the terrestrial planets. None of the terrestrial planets have rings, although Earth does have radiation belts that have been trapped.

Only Earth possesses a sizable planetary magnetic field among the terrestrials. There is no global magnetic field on Mars or the moon of Earth, although there are localised regional magnetic fields at various locations across their surfaces.

Venus, Earth, and Mars are the only terrestrial planets with noticeable atmospheres.

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Related Questions

A 1.55 kg falcon catches a 0.395 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is initially 29.5 m/s and the dove's velocity is 7.15 m/s in the same direction?

Answers

Answer:

The velocity after impact 24.96 m/s

Explanation:

Given that,

Mass of falcon = 1.55 kg

Mass of dove = 0.395 kg

Velocity of falcon = 29.5 m/s

Velocity of dove = 7.15 m/s

We need to calculate the velocity after impact

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Where,

u = initial velocity

v = final velocity

[tex]m_{1}[/tex] = mass of falcon

[tex]m_{2}[/tex] = mass of dove

Put the value in the equation (I)

[tex]1.55\times29.5+0.395\times7.15=(1.55+0.395)v[/tex]

[tex]v=\dfrac{1.55\times29.5+0.395\times7.15}{(1.55+0.395)}[/tex]

[tex]v=24.96\ m/s[/tex]

Hence, The velocity after impact 24.96 m/s

The potential difference between two points is 14.9 V when a 0.0400 C charge moves between these points by how much does its potential energy change?

Answers

Answer:

The potential energy change is 0.596 J.

Explanation:

Given that,

Potential difference =14.9 V

Charge q =0.0400 C

We need to calculate the change potential energy

The potential energy change is the product of the charge and potential difference.

Using formula of change potential energy

[tex]\Delta U=q\Delta V[/tex]

[tex]\Delta U=0.0400\times14.9[/tex]

[tex]\Delta U=0.596\ J[/tex]

Hence, The potential energy change is 0.596 J.

Answer:

0.596 J.

Explanation:

acellus

1400 kg car has a speed of 27 m/s. If it takes 7 s to stop the car, what is the Impulse and the average force acting on the car?

Answers

Explanation:

Impulse is change in momentum:

I = Δp

I = (1400 kg) (27 m/s) - (1400 kg) (0 m/s)

I = 37800 kg m/s

Impulse is also average force times time:

I = F ΔT

37800 kg m/s = F (7 s)

F = 5400 N

A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?

Answers

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120   Divide by 120

150/120 = I

5/4  = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by  √2

This question is about as sneaky as they ever get.

First let's do the easy part:

Power = (voltage) x (current)

150 watts = (120 volts) x (current)

current = (150 watts) / (120 volts)

current = 1.25 Amperes but this is NOT the answer to the question.

The voltage at the outlet is a "sinusoidal" wave ... it wiggles up and down 60 times every second.  The number of "120 volts" is NOT the "peak" of the wave.  In fact , the highest it ever gets is  √2  greater than 120 volts.  And all of this applies to the current too.

The RMS current through the lamp is (150/120) = 1.25 Amperes .

The peak current through the lamp is  1.25·√2 = about 1.77 Amperes .

How far away was a lightning strike if thunder is heard 7.20 seconds after the flash is seen? (Assume that sound traveled at 350.0m/s during the storm)

Answers

Answer: 2.52 Kilometers

Explanation:

We know that the formula to calculate speed is given by :-

[tex]\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}\\\\\Rightarrow\text{Distance}=\text{Speed}\times\text{Time}[/tex]

Given: The speed of sound during the storm : [tex]V=3350.0m/s[/tex]

The time taken by sound to travel : [tex]t= 7.20\text{ seconds}[/tex]

Then , the distance traveled by sound :-

[tex]\text{Distance}=350\times7.2=2520\text{ meters}=2.52\text{ kilometers}[/tex]

To find the distance of a lightning strike heard 7.20 seconds after the flash, multiply the sound's speed (350.0 m/s) by the time delay (7.20 s), resulting in a distance of 2520 meters.

We use the speed of sound. Since sound traveled at 350.0 m/s during the storm, we multiply the time it took to hear the thunder by the speed of sound to calculate the distance:

Distance = Speed of Sound × Time

Substituting in the given values:

Distance = 350.0 m/s × 7.20 s = 2520 meters

Therefore, the lightning struck approximately 2520 meters away.

An object moving at a constant speed of 29 m/s is making a turn with a radius of curvature of 9 m (this is the radius of the "kissing circle"). The object's momentum has a magnitude of 61 kg·m/s. What is the magnitude of the rate of change of the momentum?

Answers

Answer:

The rate of change of the momentum is 196.5 kg-m/s²

Explanation:

Given that,

Speed v = 29 m/s

Radius = 9 m

Momentum = 61 kg-m/s

We need to calculate the rate of change of the momentum

Using formula of momentum

[tex]F = \dfrac{\Delta p}{\Delta t}[/tex]

[tex]F=\dfrac{\Delta(mv)}{\Delta t}[/tex].....(I)

Using newtons second law

[tex]F = ma=m\dfrac{v^2}{r}[/tex]

[tex]F = \dfrac{mv(v)}{r}[/tex]....(II)

From equation (I) and (II)

[tex]\dfrac{\Delta p}{\Delta t}=\dfrac{61\times29}{9}[/tex]

[tex]\dfrac{\Delta p}{\Delta t}=196.5\ kg-m/s^2[/tex]

Hence, The rate of change of the momentum is 196.5 kg-m/s²

A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at a height of 2 m and has lost 25.000 J of energy to friction. The car has a mass of 800 kg. Answer the following questions. (a) How fast is the car going at time T? (b) How fast would the car be going at time T if the track were frictionless?

Answers

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

Be sure you understand why a pendulum in equilibrium hanging in a car that is accelerating forward tilts backward, and then consider the following: A helium balloon is anchored by a massless string to the floor of a car that is accelerating forward with acceleration A. Explain clearly why the balloon tends to tilt forward and find its angle of tilt in equilibrium. [Hint: Helium balloons float because of the buoyant Archimedean force, which results from a pressure gradient in the air. What is the relation between the directions of the gravitational field and the buoyant force?]

Answers

Answer:

A helium filled balloon floats forwards in a accelerating car because of the pressure difference between the front and the back of the car. When the car is accelerating, the air moves relitive to the car and the consequence is that the pressure in the back is slightly higher than in the front; which results in net force in forward direction.

hit that heart please leave brainliest and let me know if want me to answer or explain it in a different way :)

Final answer:

The helium balloon tilts forward in a car that is accelerating because the air inside the car is thrust backwards, creating more air pressure at the back and causing the balloon to float towards the lower pressure at the front. This behavior is unlike a pendulum that tilts backward due to the force of gravity.

Explanation:

The tilting of the helium balloon in a car that is accelerating forward can be explained using concepts from physics, particularly related to Newton's laws, the concept of a pendulum in equilibrium, and the buoyant Archimedean force.  

When a car accelerates, everything inside the car, including the air, is thrust backwards. This creates a higher air pressure at the back of the car than at the front. The helium balloon will float towards the lower pressure, which is at the front of the car, so the balloon will appear to tilt forward. This contrasts with the backward tilt of a pendulum, as a pendulum is influenced primarily by the force of gravity, which acts downward.

In terms of calculating the angle of tilt, it would be necessary to know the buoyant force applied to the balloon (which depends on the pressure gradient in the air), the acceleration of the car, and the mass of the balloon. However, without numerical values for these variables, a specific angle of tilt cannot be determined in this context.

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If you buy new ink cartridges for your ink-jet printer, they come in the colors "cyan", "magenta" and "yellow". Do you think that's weird? Haven't we just learned that all colors are made of "red" "green" and blue", because that's what the color receptors in the human eye can see? Why these complementary colors used?

Answers

Answer:

Let's begin by explaining that a primary color is one that can not be obtained by mixing any other color.

In this sense, red, green and blue are the primary colors of light (using the additive theory of color), but not the primary colors of the pigments.

For the case of pigments (using the subtractive color theory) the primary colors are cyan, magenta and yellow. This is because the pigments generally absorb more light than they reflect (they absorb certain wavelengths and reflect others). Therefore, the color that a given object seems to have depends on which parts of the visible electromagnetic spectrum are reflected and which parts are absorbed.

Hence, according to the subtractive theory, if we join the three primary colors of the pigments, we will obtain the black. Unlike the additive theory of light, in which if we join the three primary colors we will get white light.

Sustained current as small as 2.50×10^–6 A passing through the human heart can be fatal. Suppose a patient is undergoing open-heart surgery, and the patient’s heart has a constant resistance of 273 Ω. What is the minimum voltage across the heart that could pose a danger to the patient?

Answers

Answer:

Minimum voltage, V = 0.682 mV          

Explanation:

It is given that,

Current passing through the human heart, [tex]I=2.5\times 10^{-6}\ A[/tex]

A patient is undergoing open-heart surgery, and the patient’s heart has a constant resistance, R = 273 ohms

We have to find the minimum voltage across the heart that could pose a danger to the patient. It can be calculated using Ohm's law as :

V = IR

[tex]V=2.5\times 10^{-6}\ A\times 273\ \Omega[/tex]

V = 0.0006825 volts

or

V = 0.682 mV

Hence, the minimum voltage across the heart that could pose a danger to the patient is 0.682 mV    

A 39-kg girl is bouncing on a trampoline. During a certain interval after leaving the surface of the trampoline her kinetic energy decreases to 160 J from 490 J. How high does she rise during this interval? Neglect air resistance.

Answers

Answer:

The distance is 0.86 m.

Explanation:

Given that,

Mass = 39 kg

Initial kinetic energy, [tex]K.E_{i} = 160\ J[/tex]

Final kinetic energy, [tex]K.E_{f} = 490\ J[/tex]

We need to calculate the work done

According to work energy theorem

[tex]W = \Delta K.E[/tex]

[tex]W=K.E_{f}-K.E_{i}[/tex]...(I)

Work done is the product of the force and displacement.

[tex]W = mgh[/tex]....(II)

From equation (I) and (II)

[tex]K.E_{f}-K.E_{i}=mgh[/tex]

[tex]490-160=39\times9.8\times h[/tex]

[tex]h = 0.86\ m[/tex]

Hence, The distance is 0.86 m.

An object with a charge of −2.1 μC and a mass of 0.0044 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. (a) Find the direction and magnitude of the electric field. (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its
(1) Find the magnitude of the electric field.
(2) Find the direction of the electric field.
(3) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.
(4) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.

Answers

Answer:

(1) 2.05 x 10^4 N/C

(2) Downward

(3) upward, 9.8 m/s^2

(4) upward, 9.8 m/s^2

Explanation:

q = - 2.1 micro coulomb, m = 0.0044 kg, g = 9.8 m/s^2

(1) The electric force is given by F = - q x E

The magnitude of electric force is balanced by the weight of the charged particle

q x E = m x g

E = mg / q

[tex]E = \frac{0.0044 \times 9.8}{2.1 \times 10^{-6}}[/tex]

E = 2.05 x 10^4 N/C

(2) As the electric force is acting upward and the weight is downward so the elecric field is in downward direction.

(3) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward

(4) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward

The magnitude of the electric field and the direction of the electric filed acting on the object is,

(1) The magnitude of the electric field is [tex]2.05\times10^4\rm N/C[/tex](2) The direction of the electric field is downward.(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.What is electric field?

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

The mass of the object is 0.0044 kg, and the charge on the object is -2.1   μC.

(1) The magnitude of the electric field-

As the electric field is the ratio of force (due to gravity in given case) to the charge given on it. Thus the magnitude of the electric field of 0.0044 kg object is,

[tex]E=\dfrac{0.0044\times9.8}{2.1\times10^{-6}}\\E=2.05\times10^{4}\rm N/C[/tex]

(2) The direction of the electric field-

The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.

(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude of its acceleration-

When the electric charge on the object is doubled while its mass remains the same, then the strength of the electric field is halved. Therefore,

[tex]E=\dfrac{2.05\times10^4}{2}\\E=1.025\times10^4\rm N/C[/tex]

The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.

Hence, the magnitude of the electric field and the direction of the electric filed acting on the object is,

(1) The magnitude of the electric field is [tex]2.05\times10^4\rm N/C[/tex](2) The direction of the electric field is downward.(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.

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A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the particle's motion?Check all that apply.Increasing its chargeIncreasing its massIncreasing the field strengthIncreasing its speed

Answers

Answer:

Increasing its mass

Explanation:

More mass -> more inertia -> increased period of motion

Final answer:

Increasing the mass of the charged particle will increase the period of its motion in a uniform magnetic field. However, increasing the particle's charge, the field strength or the velocity would not result in an increased period.

Explanation:

The period of a charged particle's motion in a uniform magnetic field is determined by the equation T = 2πm/(qB), where T is the period, m is the particle's mass, q is its charge, and B is the strength of the magnetic field. One method to increase the period of motion is therefore to increase the particle's mass. Intuitively, this makes sense as a more massive particle is harder to turn, thus takes longer time to complete a cycle. However, increasing the charge q or the field strength B would decrease the period since they are in the denominator, while increasing the velocity would not affect the period in a uniform magnetic field, since the particle's speed is perpendicular to their motion.

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An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle then slows at a constant rate a2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop? Give your answers in terms of the given variables.

Answers

(a) [tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]

In the first part of the motion, the car accelerates at rate [tex]a_1[/tex], so the final velocity after a time t is:

[tex]v = u +a_1t[/tex]

Since it starts from rest,

u = 0

So the previous equation is

[tex]v= a_1 t[/tex]

So the time taken for this part of the motion is

[tex]t_1=\frac{v}{a_1}[/tex] (1)

In the second part of the motion, the car decelerates at rate [tex]a_2[/tex], until it reaches a final velocity of v2 = 0. The equation for the velocity is now

[tex]v_2 = v - a_2 t[/tex]

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

[tex]t_2=\frac{v}{a_2}[/tex] (2)

So the total time taken for the trip is

[tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]

(b) [tex]d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]

In the first part of the motion, the distance travelled by the car is

[tex]d_1 = u t_1 + \frac{1}{2}a_1 t_1^2[/tex]

Substituting u = 0 and [tex]t_1=\frac{v}{a_1}[/tex] (1), we find

[tex]d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}[/tex]

In the second part of the motion, the distance travelled is

[tex]d_2 = v t_2 - \frac{1}{2}a_2 t_2^2[/tex]

Substituting [tex]t_2=\frac{v}{a_2}[/tex] (2), we find

[tex]d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}[/tex]

So the total distance travelled is

[tex]d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]

Final answer:

The total time elapsed from start to stop for the electric vehicle is the sum of the time taken to accelerate and decelerate, calculated as T = v / a1 + v / a2. The total distance moved is the sum of the distances during acceleration and deceleration given by S = 0.5 * a1 * (v / a1)^2 + 0.5 * a2 * (v / a2)^2.

Explanation:

The question inquires about the time elapsed and the distance traveled by an electric vehicle which accelerates from rest until it reaches a certain velocity, and then decelerates to a stop. To answer part (a), we need to calculate the time taken for both acceleration and deceleration phases. For acceleration, we use the formula t1 = v / a1, and for deceleration t2 = v / a2. The total time elapsed T is then the sum of t1 and t2.

For part (b), the total distance covered S is the sum of the distances covered during acceleration and deceleration. The distance covered during acceleration s1 can be found using the equation s1 = 0.5 * a1 * t1^2, and the distance covered during deceleration s2 can be found with s2 = 0.5 * a2 * t2^2.

Hence, for an electric vehicle that accelerates to a speed v at a rate a1 and then decelerates at a rate a2 to a stop:

The total time elapsed is T = t1 + t2, which simplifies to T = v / a1 + v / a2.The total distance moved is S = s1 + s2, which simplifies to S = 0.5 * a1 * (v / a1)^2 + 0.5 * a2 * (v / a2)^2.

Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (564 nm) on a double slit with a separation of 0.108 mm. The diffraction pattern shines on the classroom wall 4.0 m away. Calculate the fringe separation between the fourth order and central fringe.

Answers

Answer:

Fringe width = 21 mm

Explanation:

Fringe width is given by the formula

[tex]\beta = \frac{\Lambda L}{d}[/tex]

here we know that

[tex]\Lambda = 564 nm[/tex]

L = 4.0 m

d = 0.108 mm

now from above formula we will have

[tex]\beta = \frac{(564 \times 10^{-9})(4.0 m)}{0.108\times 10^{-3}}[/tex]

[tex]\beta = 0.021 meter[/tex]

so fringe width on the wall will be 21 mm

Final answer:

In Young's double slit experiment with given parameters, the fringe separation between the fourth order and central fringe is calculated to be 83.6 mm.

Explanation:

The question involves calculating the fringe separation in Young's double slit experiment using a green laser with a wavelength of 564 nm, a slit separation of 0.108 mm, and a distance to the screen of 4.0 m. To find the separation between the fourth order and the central fringe, we use the formula for fringe separation in a double slit experiment, which is Δy = λL/d, where Δy is the fringe separation, λ is the wavelength of the light, L is the distance from the slits to the screen, and d is the separation between the slits. Plugging in the values, we get Δy = (564 × 10^-9 m)(4 m) / (0.108 × 10^-3 m) = 0.0209 m or 20.9 mm for the separation between each fringe. However, since we need the separation between the fourth order and the central fringe, we simply multiply by the order number, giving us 4 × 20.9 mm = 83.6 mm.

A converging lens of focal length 8.00 cm is 20.0 cm to the left of a diverging lens of focal length f-6.00 cm. A coin is placed 12.0 cm to the left of the converging lens. Find the location and the magnification of the coin's final image.

Answers

Final answer:

The location of the coin's final image is at -6.00 cm from the diverging lens, and the magnification of the image is 0.50, indicating that the image is half the size of the object.

Explanation:

To find the location and magnification of the coin's final image, we need to use the lens equation in combination with the mirror equation. First, we will calculate the position of the coin's image formed by the converging lens:

Given: Object distance (do) = -12.0 cm, Focal length (f) = 8.00 cmUsing the lens equation, 1/f = 1/do + 1/di, we can solve for the image distance (di): 1/8.00 = 1/-12.0 + 1/di. Solving for di, we get di = -24.0 cm.Since the image formed by the converging lens is on the same side as the object, it acts as an object for the diverging lens.Using the mirror equation, 1/f = 1/do + 1/di, we replace the object distance (do) with the image distance (di) from the converging lens and the focal length (f) of the diverging lens, which is -6.00 cm. Solving for the final image distance (di-final), we get di-final = -6.00 cm.

So, the location of the coin's final image is at -6.00 cm from the diverging lens. To calculate the magnification, we can use the formula m = -di/do, where di is the image distance and do is the object distance. Substituting the values, we get m = -(-6.00 cm)/(-12.0 cm) = 0.50. Therefore, the magnification of the coin's final image is 0.50, indicating that the image is half the size of the object.

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Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.1 kg attached to it and has a length of 5 m. Pendulum 2 has a ball of mass 0.9 kg attached to a string of length 1 m. How does mass of the ball affect the frequency of the pendulum?

Answers

Explanation:

For pendulum 1 :

Mass, m₁ = 0.1 kg

Length, l₁ = 5 m

For pendulum 2 :

Mass, m₂ = 0.9 kg

Length, l₂ = 1 m

The time period the simple pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]..........(1)

And we know that frequency f is given as :

[tex]f=\dfrac{1}{T}[/tex]

[tex]f=\dfrac{1}{2\pi\sqrt{\dfrac{L}{g}}}[/tex]

From equation (1) it is clear that the time period and frequency depend on the length of the bob and acceleration due to gravity only. It is independent of its mass.

A satellite weighs 104 N at ground control. What best approximates the acceleration it experiences in orbit at an altitude of twice the earth's radius ifFpa = GME2thm, where r is the distance separating the centers of mass of the satellite and the Earth? O A. 111 m/s2 O B. 2.5 m/s O C. 1.1 m/s2 ○ D. 0 m/s

Answers

Answer:

[tex]a_c = 1.1 m/s^2[/tex]

Explanation:

As we know that net force on the Satellite due to gravity will provide it centripetal force

so we can say here

[tex]F_g = F_c[/tex]

[tex]\frac{GMm}{r^2} = ma_c[/tex]

now we will have

acceleration given by the equation

[tex]a_c = \frac{GM}{r^2}[/tex]

now we have

r = R + 2R = 3R

[tex]a_c = \frac{GM}{9R^2}[/tex]

also we know that acceleration due to gravity on the surface of earth is given as

[tex]g = \frac{GM}{R^2} = 9.8 m/s^2[/tex]

so the acceleration of satellite is given as

[tex]a_c = \frac{9.8}{9} = 1.1 m/s^2[/tex]

Final answer:

The acceleration of a satellite in orbit at an altitude of twice Earth's radius is best approximated using the formula for gravitational acceleration, g = GME / r². Plugging in the values and simplifying, the acceleration is approximately 2.45 m/s², with option B (2.5 m/s²) being the closest approximation.

Explanation:

Calculating the Acceleration of a Satellite in Orbit

In order to best approximate the acceleration a satellite experiences in orbit at an altitude of twice the Earth's radius, we must apply Newton's law of universal gravitation and the formula for gravitational acceleration, g = GME / r². Given that the satellite weighs 104 N at ground control which reflects the gravitational force at Earth's surface, we use this formula to find its gravitational acceleration in orbit. The relevant equation provided, Fpa = GME2thm/r, appears to be a typo, but our main equation for gravitational acceleration does not require mass, as it cancels out during the calculation:

g = GME / r²

Since we're given that the altitude is twice the Earth's radius, the distance r from Earth's center to the satellite in orbit is 3RE (1RE for Earth's radius and 2RE for the altitude above Earth). Considering this, we can calculate:

g = (6.674 x 10-11 m³/kg s²) (5.972 x 1024 kg) / (3 x 6.371 x 106 m)²

After simplifying, we get an acceleration of approximately 2.45 m/s². Therefore, option B, 2.5 m/s², best approximates the acceleration the satellite experiences in orbit at this altitude.

a mass suspended by a spring stretches an additional 4 ccm when an additional 10 gram mass is attached to it. what is the value of the spring constant k, in si units

Answers

Answer:

2.45 N/m

Explanation:

We can find the spring constant by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force applied to the spring

k is the spring constant

x is the stretching of the spring

Here, the force applied is the weight of the mass hanged on the spring. The mass is

m = 10 g = 0.010 kg

So the weight is

[tex]F=mg=(0.010 kg)(9.8 m/s^2)=0.098 N[/tex]

while the stretching is

x = 4 cm = 0.04 m

So the spring constant is

[tex]k=\frac{F}{x}=\frac{0.098 N}{0.04 m}=2.45 N/m[/tex]

A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 2.43 s?

Answers

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{4.86\ m/s-0}{2.43\ s}[/tex]

a₁ = 2 m/s²

(b) Acceleration of the car,

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}[/tex]

a₂ = 4.97 m/s²

(c) Distance covered by the car,

[tex]d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2[/tex]

[tex]d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2[/tex]

d₁ = 5.904 m

Distance covered by the jogger,

[tex]d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2[/tex]

[tex]d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2[/tex]

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the person and the kayak is 74 kg, what is the magnitude of the net force acting on the kayak?

Answers

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

Final answer:

The magnitude of the net force acting on the kayak, which accelerates from 0 to 0.61 m/s over a distance of 0.39 m with a mass of 74 kg, is approximately 35.25 Newtons (N).

Explanation:

A person in a kayak accelerates from 0 to 0.61 m/s over a distance of 0.39 meters. To find the magnitude of the net force acting on the kayak, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. We can calculate this as:

Kinetic energy (KE) = (1/2)mv2Work done (W) = Force (F) × Distance (d)So, the net work done is also equal to the change in kinetic energy, KE = W

To find the net force, we need to first calculate the change in kinetic energy, which is the kinetic energy at 0.61 m/s minus the initial kinetic energy at 0 m/s:

ΔKE = (1/2)m(v2 - 02)ΔKE = (1/2) × 74 kg × (0.61 m/s)2ΔKE = (1/2) × 74 kg × 0.3721 m2/s2ΔKE = 13.7467 kg·m2/s2 (= 13.7467 Joules)

Since work done is equal to force times distance and to the change in kinetic energy, we can express the net force as:

F = ΔKE / dF = 13.7467 J / 0.39 mF = 35.247 N  (rounded to three decimal places)

Therefore, the magnitude of the net force acting on the kayak is approximately 35.25 Newtons (N).

A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.

Answers

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

[tex]\frac{V_1}{N_1}=\frac{V_2}{N_2}[/tex]

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

[tex]V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V[/tex]

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

A car is traveling at 53.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.104, what is the minimum distance in which the car will stop?

Answers

Answer:

902 ft

Explanation:

First convert mi/h to ft/s.

53.0 mi/h × (5280 ft / mi) × (1 h / 3600 s) = 77.7 ft/s

Sum of the forces on the car in the y direction:

∑F = ma

N - W = 0

N = mg

Sum of the forces on the car in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substituting;

-mgμ = ma

-gμ = a

Acceleration is constant, so:

v² = v₀² + 2a(x - x₀)

(0 ft/s)² = (77.7 ft/s)² + 2(-32.2 ft/s² × 0.104)(x - 0)

x = 902 ft

The minimum stopping distance is 902 ft.

A jet airplane lands with a speed of 120 mph. It has 1800 ft of runway after touch- down to reduce its speed to 30 mph. Compute the average acceleration required of the airplane during braking A: a -8.1 ft/s2

Answers

Answer:

The average acceleration is 8.06 m/s².

Explanation:

It is given that,

Initial speed of the jet, u = 120 mph = 176 ft/s

Final velocity of the jet, v = 30 mph = 44 ft/s

Distance, d = 1800 ft

We need to find the average acceleration required of the airplane during braking. It can be calculated using third law of motion as :

[tex]v^2-u^2=2ad[/tex]

a = acceleration

[tex]a=\dfrac{v^2-u^2}{2d}[/tex]

[tex]a=\dfrac{(44\ ft/s)^2-(176\ ft/s)^2}{2\times 1800\ ft}[/tex]

[tex]a=-8.06\ ft/s^2[/tex]

So, the average acceleration required of the airplane during braking is -8.06 ft/s². Hence, this is the required solution.

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the projectile to reach the ground below. (b) Determine the distance from the base of the building that the projectile lands. (c) Determine the horizontal and vertical components of the velocity just before the projectile reaches the ground.

Answers

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

[tex]y = h + \frac{1}{2}at^2[/tex]

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

[tex]0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s[/tex]

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

[tex]v_x = 23.4 m/s[/tex]

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

[tex]d=v_x t = (23.4 m/s)(3.35 s)=78.4 m[/tex]

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

[tex]v_x = 23.4 m/s[/tex]

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

[tex]v_y = u_y +at[/tex]

where

[tex]u_y = 0[/tex] is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

[tex]v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s[/tex]

and the negative sign means it points downward.

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

(a) The time necessary for the projectile to reach the ground below is 3.35 seconds.(b) The distance from the base of the building that the projectile lands is 78.4 meters.(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

For the motion of horizontal and vertical direction we use following equation in projectile motion.

[tex]y=y_o+\dfrac{1}{2}gt^2[/tex]

Here, (g) is the gravity and (t) is time.

Here, in the given problem, the projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall.

(a) The time necessary for the projectile to reach the ground below-

As the height of the projectile before the shot is 55 meters and the the height of the projectile is 0 meter when it reaches to the ground. Thus from the above equation,

[tex]0=55+\dfrac{1}{2}\times9.81\times(t)^2\\t=3.35\rm s[/tex]

Thus the time required for the projectile to reach the ground below is 3.35 seconds.

(b) The distance from the base of the building that the projectile lands-

The distance traveled by the object is the product of velocity and time taken. Thus, distance from the base of the building that the projectile lands is,

[tex]d=23.4\times3.35\\d=78.4\rm m[/tex]

(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground-

Horizontal component of velocity is equal to the velocity of projectile shot just before it reaches to the ground. Thus, horizontal component of velocity is,

[tex]v_x=23.4\rm m/s[/tex]

Now the time taken is 3.35 seconds. Thus the vertical component of velocity can be found with the formula of first equation of motion with  0 initial velocity as,

[tex]v_y=0+(-9.81)\times(3.35)\\v_y=-32.8\rm m/s[/tex]

The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.

(a) The time necessary for the projectile to reach the ground below is 3.35 seconds.(b) The distance from the base of the building that the projectile lands is 78.4 meters.(c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.

Learn more about the projectile motion here;

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An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C.225 ∘C. During this process, 346 J of work was done by the system and its internal energy increased by 6085 J.6085 J. What is the specific heat of the gas?

Answers

Final answer:

The specific heat of the gas is 0.381 J/g°C.

Explanation:

The specific heat of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. To find the specific heat of the gas, we can use the equation:



Q = mass * specific heat * change in temperature



In this case, the mass of the gas is 80.0g, the change in temperature is 225°C - 25°C = 200°C, and the energy absorbed by the gas is 6085J. Plugging these values into the equation, we can solve for the specific heat:



6085J = 80.0g * specific heat * 200°C



Specific heat = 6085J / (80.0g * 200°C) = 0.381 J/g°C

Learn more about specific heat here:

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Final answer:

The specific heat of the given gas can be calculated using the formula c = Q / (m * ΔT). By substitifying Q = 6431J (sum of work done by the system and increase in internal energy), m = 80g (mass of gas), and ΔT = 200°C (change in temperature), we find that the specific heat of the gas is roughly 0.40 J/g°C

Explanation:

The specific heat of a substance is the heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For the given gas, we first need to calculate the heat absorbed which is the sum of the work done by the system and the increase in its internal energy, which equals 346J + 6085J = 6431J. The following information is given, mass m = 80.0g and the change in temperature ΔT = 200°C (225 °C - 25°C).

The formula for specific heat is:
Q = m*c*ΔT
where:
Q is the heat energy absorbed (or released),
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
Solving this equation for the specific heat (c), we find:
c = Q / (m * ΔT).

So, substituting the given values into this formula, we have:
c = 6431J /  (80.0g * 200°C = 0.40 J/g °C which will be the specific heat of the gas.

Learn more about Specific Heat here:

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If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.

Answers

When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

An object is said to be in a condition of rotational equilibrium if there is no net torque operating on it. In other words, the object is keeping a constant angular velocity while not suffering any rotational acceleration. The object has a constant angular momentum when the net torque is zero.

If the net torque is zero, the item is being pulled in a direction that is equal to the sum of all the clockwise and anticlockwise torques pulling on it. To put it another way, any propensity the object has to spin in one direction is counterbalanced by an equal tendency to rotate in the opposing direction.

Hence, When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.

To learn more about Torque, here:

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Final answer:

When the net torque is zero, it means that the clockwise and counterclockwise torques are balanced.

Explanation:

When the net torque is zero, it implies that the clockwise and counterclockwise torques balance each other out. In other words, the total torque in the clockwise direction is equal in magnitude but opposite in sign to the total torque in the counterclockwise direction.

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?

Answers

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

[tex]F \Delta t = m (v-u)[/tex]

where

F is the average force

[tex]\Delta t[/tex] is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

[tex]\Delta t = 60.0 ms = 0.06 s[/tex]

Solving for F,

[tex]F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N[/tex]

And since we are interested in the magnitude only,

F = 106.7 N

The magnitude of the average force applied to the ball was about 107 N

[tex]\texttt{ }[/tex]

Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial velocity of the ball = u = -20.0 m/s

final velocity of the ball = v = 12.0 m/s

contact time = t = 60.0 ms = 0.06 s

mass of the ball = m = 0.200 kg

Asked:

average force applied to the ball = F = ?

Solution:

We will use this following formula to solve this problem:

[tex]\Sigma F = ma[/tex]

[tex]F = m ( v - u ) \div t[/tex]

[tex]F = 0.200 ( 12 - (-20) ) \div 0.06[/tex]

[tex]F = 0.200 ( 32 ) \div 0.06[/tex]

[tex]F = 6.4 \div 0.06[/tex]

[tex]F = 106 \frac{2}{3} \texttt{ N}[/tex]

[tex]F \approx 107 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441Newton's Law of Motion: https://brainly.com/question/10431582Example of Newton's Law: https://brainly.com/question/498822

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Consider a solid sphere of radius R = 0.4 m that is uniformly charged with ? = -11 ?C/m3. What is the electric potential a distance 5 m from the center of the sphere?

Answers

Answer:

V=-5304.6V

Explanation:

(I set the charge density unit to microcoulombs per metre cubed)

Answer:

In the picture.

Final answer:

To find the electric potential 5 m from the center of the uniformly charged sphere, the sphere is treated as a point charge resulting from its total charge due to the volume charge density. The potential is calculated using Coulomb's law for a point charge, yielding a value of -0.399 MV.

Explanation:

To calculate the electric potential a distance 5 m from the center of a uniformly charged solid sphere with radius R = 0.4 m and charge density [tex]\\rho = -11 \mu C/m^3\[/tex], we use the concept of electric potential due to a continuous charge distribution. Since the point where we need to find the potential is outside the sphere, we can treat the sphere as a point charge located at its center with total charge Q.

The total charge Q can be found by integrating the charge density over the volume of the sphere:

[tex]Q = \int \rho \, dV = \rho \frac{4}{3}\pi R^3[/tex]

Plugging the values in, we have:

[tex]Q = -11 \times 10^{-6} C/m^3 \times \frac{4}{3}\pi \times (0.4 m)^3\\Q = -2.21 \times 10^{-7} C[/tex]

Now, the electric potential V at a distance r from a point charge Q is given by:

[tex]V = \frac{kQ}{r}[/tex]

where k is Coulomb's constant, k = 8.99 \times 10^9 N m^2/C^2 and r is the distance from the center of the sphere, which is 5 m in this case. Thus:

[tex]V = \frac{8.99 \times 10^9 N m^2/C^2 \times (-2.21 \times 10^{-7} C)}{5 m}[/tex]

V = -0.399 MV

Molecules cool a cloud and are usually located on the edges of the cloud. T/F

Answers

I think true but not sure about it
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