Help me pleaseee!!
Would appreciate ^^
**BRAINLIST, thanks and
points***

The battery has an internal resistance of 2.50
4.1 Show that the resistance of the 6.2 V, 4.5 W lamp at its working potential difference
(pd) is about 9.

Look at photo for more info
4.2 The terminal pd across the battery is 6.2 V.
Calculate the emf of the battery.

rope
Pulleys P and M are frictionless so that the tension in the rope attached to A is equal to the weight of A
A weighs 35 N and the weight of moveable pulley M is negligible.

Calculate the tension in the horizontal cable C when the gate is closed.
tension
3.3 Pulley M is pulled to the left as the gate is opened.
10/36
Explain why this increases the tension in the horizontal cable C.

Help Me Pleaseee!!Would Appreciate ^^**BRAINLIST, Thanks Andpoints***The Battery Has An Internal Resistance

Answers

Answer 1

The resistance of working potential difference of the 6.2 V, 4.5 W lamp is 8.54 ohm. When terminal pd across the battery is 6.2 V, the emf of the battery is 8.01 V.

What is resistance?

The opposition that a substance provides to the flow of electric current is referred to as resistance. The uppercase letter R represents it. The ohm is the standard unit of resistance, which is sometimes written as a word and sometimes represented by the uppercase Greek letter omega. Resistance is a measure of an electrical circuit's resistance to current flow. Resistance is measured in ohms, which is represented by the Greek letter omega ().

Here,

r=2.5 ohm

V= 6.2 volts

P=4.5 watt

P=V²/R

R=V²/P

R=6.2*6.2/4.5

R=8.54 ohm that is about 9 ohm.

I=V/R

I=0.726 A

EMF=V+Ir

=6.2+0.726*2.5

=6.20+1.81

=8.01 volts

The resistance of the 6.2 V, 4.5 W lamp at its working potential difference is 8.54 ohm. The emf of the battery when terminal pd across the battery is 6.2 V is 8.01 V.

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Related Questions

A photon of wavelength 29 pm is scattered by a stationary electron. What is the maximum possible energy loss of the photon? (m el = 9.11 × 10-31 kg, h = 6.626 × 10-34 J · s, c = 3.00 × 108 m/s)

Answers

The maximum possible energy loss of the photon is 2.7 x 10-14 J, which is much less than the energy of the photon, 7.3 x 10-19 J.

To Calculate the energy of the photon.

E = hc/λ

E = (6.626 × 10-34 J · s)(3.00 × 108 m/s)/(29 x 10-12 m)

E = 7.3 x 10-19 J

To Calculate the maximum possible energy loss of the photon.

Max energy loss = 2m el (c2)

Max energy loss = 2(9.11 x 10-31 kg)((3.00 x 108 m/s)2)

Max energy loss = 2.7 x 10-14 J

Compare the energy of the photon and the maximum possible energy loss.

Hence, maximum possible energy loss of the photon is 2.7 x 10-14 J, which is much less than the energy of the photon, 7.3 x 10-19 J.

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If the number of turns on the secondary coil of a transformer are less than those on the primary, the result is a.
A) step-up transoformer.
B) a dc transformer.
C) 120-v transformer
D) step-down transoformer
E) 220-v transformer

Answers

im pretty sure it’s C

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s= 2 sin π t + 3 cos π t, where t is measured in seconds.
a. Find the average velocity during each time period:
i (1, 2)
ii. (1, 1.1)
iii. (1, 1.01)
iv. (1, 1.001)
b. Estimate the instantaneous velocity of the particle when t=1.

Answers

(a.i) The average velocity of the particle for period (1, 2) is 0 cm/s.

(a.ii) The average velocity of the particle for period (1, 1.1) is - 3.1 cm/s.

(a.iii) The average velocity of the particle for period (1, 1.01) is -3.3 cm/s.

(a.iv) The average velocity of the particle for period (1, 1.001) is -3 cm/s.

(b) The instantaneous velocity of the particle when t=1 is - 2 cm/s.

What is average velocity?

The average velocity of an object is the ratio of the total displacement to total time of motion of the object.

v = ( total displacement ) / ( total time )

v = ( s ) / ( t )

The average velocity of the particle for period (1, 2)

s = 2 sin π (1) + 3 cos π (1)       +      2 sin π (2)  +  3 cos π (2)

s = ( 0 - 3 ) + ( 0 + 3 ) = 0 cm

v = ( 0 cm ) / (1 + 2 ) = 0 cm/s

The average velocity of the particle for period (1, 1.1)

s = 2 sin π (1) + 3 cos π (1)       +      2 sin π (1.1)  +  3 cos π (1.1)

s = ( 0 - 3 ) + ( -0.62  - 2.85 ) = -6.47 cm

v = ( -6.47 cm ) / ( 1 + 1.1 )

v = -3.1 cm/s

The average velocity of the particle for period (1, 1.01)

s = 2 sin π (1) + 3 cos π (1)       +      2 sin π (1.01)  +  3 cos π (1.01)

s = ( 0 - 3 ) + ( -0.63  - 3 ) = -6.63 cm

v = ( -6.63 cm ) / ( 1 + 1.01 )

v = -3.3 cm/s

The average velocity of the particle for period (1, 1.001)

s = 2 sin π (1)  + 3 cos π (1)       +      2 sin π (1.001)  +  3 cos π (1.001)

s = ( 0 - 3 ) + ( -0.0063  - 3 ) = -6.0063 cm

v = ( -6.0063 cm ) / ( 1 + 1.001 )

v = -3 cm/s

The instantaneous velocity of the particle when the time = 1, is calculated as follows;

v (t) = ds / dt

v (t) = 2 cosπ t  - 3 sin π t

v (1) = 2 cosπ (1)  - 3 sin π (1)

v ( 1 ) = -2 cm /s

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Force (N)
896760 50 3NTO
10
2
I
0 0.04 0.08 0.12 0.16 0.20
Extension (m)
13. Using the graph above, (a) determine the spring constant, (b) write the Hooke's law equation for this
spring.

Answers

a) The spring constant is 896760 N/m.

b) F = -896760x

How do we arrive at the values?

(a) To determine the spring constant, we need to find the slope of the linear portion of the graph. The slope can be found by dividing the change in force by the change in extension.

(b) Hooke's law states that the force applied to a spring is proportional to the spring's extension, with the constant of proportionality being the spring constant. So the equation for this spring would be:

F = kx

where F is the force, x is the extension, and k is the spring constant.

Therefore, the spring constant is 896760 N/m. It could then be concluded that the spring constant is as given above.

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In which of the following situations is the velocity constant, but the acceleration is not zero? In no situation can velocity be constant at the same time there is non-zero acceleration A car drives at constant speed along a winding road. A sprinter runs a 100-meter dash along a straight track.A train moves along a straight track at constant speed

Answers

In no case can velocity be constant while there is non-zero acceleration.

What is the relation between velocity and acceleration?During uniformly accelerated, straight-line motion, the relationship between velocity and time is straightforward. The greater the change in velocity, the longer the acceleration. When the acceleration is constant, the change in velocity is directly proportional to time.This is also the case in uniform motion, where the object moves at a constant speed. Because acceleration implies a non-zero change in velocity with respect to time, a particle can have zero acceleration if the velocity is constant but non-zero.When the magnitude and direction of a velocity do not change over time, it is said to be constant. In other words, this is when the rate of change of an object's position remains constant over time.

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consider the following observations. classify each observation based on whether it is a real observation (a true statement of something we can actually see from earth) or one that is not real (a statement of something that does not really occur as seen from earth). view available hint(s)for part a

Answers

Real: True statements

Mercury undergoes a complete cycle of phases.Daily moon rises in the east and sets in the west.Stars circle daily around north or south celestial poleEach year, neighbouring stars' positions gradually oscillate back and forth.A distance galaxy rises in east, sets in west each day

Not real: False statements:

We sometimes see a crescent JupiterBeyond Saturn, a planet rises in the west and sets in the east

Earth observation is the gathering of information on the physical, chemical, and biological processes occurring on planet Earth using remote sensing technology, often via satellites carrying imaging equipment. Earth observation is used to monitor and assess environmental changes, both natural and man-made, as well as their state.

Seismic and Global Positioning System (GPS) stations, as well as floatable buoys for monitoring ocean currents, temperature, and salinity, are some of the techniques used today for Earth observation. Air quality and rainfall patterns are also recorded by land-based stations. Space-based technologies provide repeatable datasets that provide a unique opportunity to gather data about the world when combined with appropriate technique development and analysis.

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Following question may be like this:

Consider the following observations. Classify each observation based on whether it is a real observation (a true statement of something we can actually see from Earth) or one that is not real (a statement of something that does not really occur as seen from Earth).

- Mercury goes through a full cycle of phases

- Moon rises in east, sets in west each day

- stars circle daily around north or south celestial pole

- positions of nearby stars shift slightly back and forth each year

- a distance galaxy rises in east, sets in west each day

- we sometimes see a crescent Jupiter

- a planet beyond Saturn rises in west, sets in east

A helicopter flies horizontally at a contant speed. This makes a 42500 N lift force at 78.3 degrees, and air resistance pushes back against it. What is the mass of the helicopter?

Answers

In order to calculate the mass of the helicopter, we must first calculate the force of the air resistance.

What is the mass of the helicopter?

The mass of the helicopter can be calculated using the formula:

mass = lift force / (acceleration due to gravity x sin(angle of lift force))

mass = 42500 N / (9.81 m/s2 x sin(78.3°))

mass = 4289 kg.

The force of the air resistance is equal to the magnitude of the lift force multiplied by the sine of the angle of the lift force, which in this case is 42500N x sin(78.3°) = 39,941N. Now, since the lift force and the air resistance are equal and opposite, we can use Newton's second law of motion, F=ma, to calculate the mass of the helicopter.Thus, the mass of the helicopter is equal to the force of the air resistance divided by the acceleration of the helicopter, which is zero since it is moving at a constant speed.Therefore, the mass of the helicopter is 39,941N/0m/s2 = 39,941kg.The mass of the helicopter can be calculated using the principles of Newton's Second Law of Motion. According to this law, the force acting on an object is equal to its mass times its acceleration. Therefore, the mass of the helicopter can be calculated using the equation:

m = F / a

where F is the lift force and a is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, the mass of the helicopter can be calculated as:

m = 42500 N / (9.81 m/s^2)

m = 4317.3 kg

This theory is known as Newton's Second Law of Motion, which states that the net force on an object is equal to its mass times its acceleration. This law can be used to calculate the mass of the helicopter when the lift force and acceleration due to gravity are known.

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The resultant of a 40-N force at right angles to a 30-N force is

Answers

Answer:

The magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.

Explanation:

The resultant of two forces at right angles to each other is found by using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the longest side, opposite the right angle) is equal to the sum of the squares of the other two sides (the two shorter sides).

In this case, the two forces form the two shorter sides of the triangle and the resultant force is the hypotenuse.

So the magnitude of the resultant force can be calculated as:

√(40^2 + 30^2) = √(1600 + 900) = √2500 = 50 N

The magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.


The speed of light is about 300,000 km/sec. The average distance between the Earth and the Sun (1
AU) is about 150 million km. Approximately how long will it take light to travel to 1 AU?

Answers

It will take approximately 8.3 minutes for light to travel from the Earth to 1 AU

The speed of light is a constant value of about 300,000 km/sec. To find out how long it will take light to travel to 1 AU (which is about 150 million km), we can use the formula:

Time = distance/speed

So we can substitute the values into the formula:

time = 150 million km / 300,000 km/sec

By solving the equation we get:

time = 500 seconds or approximately 8.3 minutes

Therefore, it will take approximately 8.3 minutes for light to travel from the Earth to 1 AU, which is the average distance between the Earth and the Sun.

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What are 2 contradictions of inertia

Answers

A body's inertia is a passive characteristic that only allows it to oppose active agents like forces and torques. A moving body continues to move not because of its inertia but rather because there is no external force present to cause it to slow down, veer off course, or accelerate.

What is inertia?

The idea of inertia states that an object will keep moving in the same direction until another force causes it to change its direction or speed.

It is important to recognise that Newton used the slang phrase "the principle of inertia" when he discussed it in his first law of motion.

resistance of the body to changes in momentum

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the velocity of a 1.7 kg block sliding down a frictionless inclined plane is found to be 1.54 m/s. 1.70 s later, it has a velocity of 6.13 m/s.

Answers

The angle of plane is 15 degrees.

mg x Sin A gives the force that pushes the mass down the incline.

Additionally, this force uses the same formula m x a which is mass times acceleration,

where a is the original acceleration in the plane.

The block sliding has a mass of 1.7 kg and an initial speed of 1.54 m/s.

the velocity is 6.13 m/s after the time of 1.70 seconds, which is given.

By dividing the difference between the two initial velocities by the given time, or

(6.13 m/s - 1.54 m/s) / 1.70s = 2.70 [tex]m/s^2[/tex],

the acceleration a may be calculated.

ma is therefore which is  equal to mass time acceleration g times sin A, or

sin A = acceleration (a)  / gravity (g) , or

 2.70 [tex]m/s^2[/tex] times 9.81  times 0.275.

A is equal to arc sin (0.275) = or sin A  = 15 degrees

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A stone is thrown downwards with a speed of 5ms, from the top of a building of height 180 m. Take point of projection as the origin and vertical downward direction as negative. Determine
a. the velocity of stone at the end of 3 s.
b. the displacement of stone at the end of 3 s.
c. the time taken by the stone to travel a distance of 150 m.

Answers

Explanation:

open the image for detailed solution

Why is B the correct answer? Thanks :)

Answers

The final speed is v and the change in the total kinetic energy is 6mv² (Option B)

How do I determine the final velocity?

The final speed after the collision can be obtained as illustrated below:

Mass of 1st object (m₁) = 2mspeed of 1st object (u₁) = 3vMass of 2nd object (m₂) = 4mspeed of 2nd object (u₂) = 0Final speed (v) = ?

Momentum before = momentum after

m₁u₁ + m₂u₂ = v₀(m₁ + m₂)

Divide both sides by (m₁ + m₂)

v₀ = [m₁u₁ + m₂u₂] / (m₁ + m₂)

v₀ = [(2m × 3v) + (4m × 0)] / (2m + 4m)

v₀ = [6mv + 0] / 6m

v₀ = 6mv / 6m

v₀ = v

Thus, the final speed is v

How do I determine the change in the kinetic energy?

First, we shall determine the total initial kinetic energy. Details below:

Mass of 1st object (m₁) = 2mspeed of 1st object (u₁) = 3vMass of 2nd object (m₂) = 4mspeed of 2nd object (u₂) = 0Total Initial Kinetic energy (KE₁) =?

KE₁ = ½m₁u₁² + ½m₂u₂²

KE₁ = [½ × 2m × (3v)²] + [½ × 2m × 0²]

KE₁ = [m × 9v²] + 0

KE₁ = 9mv²

Next, we shall determine the total final kinetic energy. Details below:

Total mass (m) = 2m + 4m = 6mFinal speed (v₀) = vTotal final Kinetic energy (KE₂) =?

KE₂ = ½mv²

KE₂ = ½ × 6m × v²

KE₂ = 3mv²

Finally, we shall determine the change in the kinetic energy of the car can be obtained as follow:

Total Initial Kinetic energy (KE₁) = 9mv²

Total final Kinetic energy (KE₂) = 3mv²

Change in total kinetic energy (ΔKE) =?

ΔKE =KE₁ - KE₂

ΔKE = 9mv² - 3mv²

ΔKE = 6mv²

In conclusion,

Final speed = vChange in total kinetic energy = 6mv²

Thus, the correct answer is Option B

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Calculate the first and second velocities of the car with one washer attached to the pulley, using the formulas

v1 = 0.25 m / t1, and
v2 = 0.25 m / (t2 – t1)
where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, rounded to two decimal places, in Table E.

What is the first velocity of the car with one washer at the 0.25 meter mark?

m/s

What is the second velocity of the car with one washer at the 0.50 meter mark?

m/s

Answers

The speed is unusual (65 [tex]\frac{mi}{h}[/tex] + 55[tex]\frac{mi}{h}[/tex] ), The two rates were not maintained keep up with the times.

Calculating the problem:

The speed changed once a certain amount of time had passed. Speed = We get 65 = [tex]\frac{mi}{h}[/tex] = [tex]\frac{130 mi}{h}[/tex] by substituting variables, which is the formula for this = [tex]\frac{d}{t}[/tex]

In the interim, t1 has a value of 2.0 hours.

At lower speeds, the time is shown as t2 = T - t1 = 3.33 h — 2.0 h = 1.33 h.

The difference between the two speeds, such as speed = 73 miles, is also shown in this.

D is the sum of d1 and d2, which equals 203 miles.

The normal speed is = [tex]\frac{d}{t}[/tex] = [tex]\frac{203mi}{h}[/tex]= 61.

The two rates were not kept up with the times because of the unusual speed (65[tex]\frac{mi}{h}[/tex] + 55 [tex]\frac{mi}{h}[/tex]).

What is speed?

The direction in which a body or object is moving is determined by velocity.

Most of the time, velocity is a scalar number. In essence, velocity is a vector quantity. The term "velocity" refers to the rate of change in displacement and distance. speed is indicated by the speed and direction of travel (for instance, 60 kilometers per hour north).

Typically, velocity is defined as the vector value of the speed and motion direction. The speed at which something moves in one direction is known as velocity. Velocity can be used to measure both the speed of a car traveling north on a highway and the speed of a rocket taking off into space.

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Answer:

What is the first velocity of the car with one washer at the 0.25 meter mark?

1st  = 0.11 m/s

2nd = 0.13 m/s

3rd = 0.19 m/s

What is the second velocity of the car with one washer at the 0.50 meter mark?

1st = 0.28 m/s

2nd = 0.36 m/s

3rd = 0.45 m/s

Explanation:

trust

(a) what is the velocity vpi of the plank relative to the surface of the ice? (use the following as necessary: vgp, mg, and mp. indicate the direction with the sign of your answer. let the positive direction be in the direction that the girl walks.)

Answers

The final velocity of plank with respect to the ice will be (-Mg×Vgp)/(Mg+Mp) and it will be in direction opposite to the girl.

Mass of the girl, = Mg

Mass of the plank, = Mp

Velocity of the girl with respect to plank, = Vgp

Velocity of the plank with respect to ice, = Vpi

Velocity of girl with respect to ice, = Vgi

Initially both girl and the plank were at rest with respect to ice, so initial momentum of both was zero. So final momentum of both with respect to ice will be zero.

Mg×Vgi + Mp×Vpi = 0

Mg×Vgi = -Mp×Vpi .....equation (1)

We know that velocity of girl with respect to the ice will be the sum of velocity of the girl with respect to the plank and the velocity of the plank with respect to the ice.

Vgi = Vpi + Vgp.... put this value in equation (1)

Mg×(Vpi + Vgp) = -Mp×Vpi

Mg×Vpi + Mg×Vgp) = -Mp×Vpi

Vpi(Mg + Mp) = -Mg×Vgp

Vpi = -Mg×Vgp/(Mg + Mp)

A negative sign indicates that the velocity of the plank will be in the opposite direction to the girl.

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--The give question is incomplete, the complete question is:

"A girl of mass Mg is standing on a plank of mass Mp. Both are originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity Vgp to the right relative to the plank. (The subscript gp denotes the girl relative to the plank.)

What is the velocity Vpi of the plank relative to the surface of the ice? (Use the following as necessary: Vgp, Mg, and Mp. Indicate the direction with the sign of your answer. Let the positive direction be in the direction that the girl walks.)"--

as infants extract an increasing number of potential word forms from the speech stream they hear, they begin to associate these with concrete, perceptually available objects in their world. this is referred to as multiple choice question.

Answers

Infants extracting an increasing number of potential word forms from the speech stream they hear, begin to associate these with concrete, perceptually available objects is referred to as statistical learning.

Statistical learning refers to the process by which infants extract potential word forms from the speech stream they hear and begin to associate them with concrete, perceptually available objects. This process is thought to be a crucial component of language development, as it allows infants to begin to understand the structure and meaning of the language they are exposed to.

Through statistical learning, infants are able to identify patterns and regularities in the speech they hear, such as the relationship between a word and the object it refers to. This ability enables them to begin to build a vocabulary and develop an understanding of the grammar and syntax of their native language. Additionally, statistical learning may also play a role in the development of other cognitive abilities, such as memory and attention.

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Your question seems incomplete, but I suppose the question was:

"As infants extract an increasing number of potential word forms from the speech stream they hear, they begin to associate these with concrete, perceptually available objects in their world. this is referred to as"

a particle of uniform mass m is locaed inside a uniform solid sphere of radius r and mass m at a distance r from its center (a) sho that the gravitational potential energy of the system is U=GMmr^2 /2R^3 - 3GmM/2R ?

Answers

(a) The gravitational potential energy of the system is  [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] (Proved), and (b)  the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center will [tex]-\frac{GMm}{2R}[/tex].

Assuming that a concentric spherical surface passes through where a particle of mass m resides.

so from that imaginary boundary

The mass of the inner part is

M' = [tex]\frac{M}{\frac{4\pi r^3}{3} } . \frac{4}{3} \pi r^3[/tex]= Mr³/r³

The potential at P due to this inner part is

V₁ = -GM'/r

or V₁ = -GMr²/R³

Now, to find the potential at P due to the outer part of the sphere, we will divide this by concentric shells.

so, dm =  [tex]\frac{M}{\frac{4\pi r^3}{3} } . \frac{4}{3} \pi x^2 dx[/tex] = 3Mx²dx/R³

by further solving,

[tex]-\frac{Gdm}{x} = -3\frac{GM}{R^3}xdx[/tex]

so by integrating,

V₂ = [tex]\int\limits^R_r -3\frac{GM}{R^3}xdx[/tex]

or V₂ = [tex]-\frac{3GM}{2R^3} (R^2-r^2)[/tex]

So, the total potential at P will be

V = V₁ + V₂ = -GMr²/R³  [tex]-\frac{3GM}{2R^3} (R^2-r^2)[/tex]

or V = [tex]-\frac{GM}{2R^3}(3R^2-r^2)[/tex]

and we know that potential energy U = mV = [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] {Hence Proved}

(b) we know that potential energy at the center is Uc = [tex]-\frac{3GmM}{2R}[/tex] {r=0}

At surface {r+R} Us = [tex]-\frac{GMm}{R}[/tex]

hence, work done W = Uc - Us = [tex]-\frac{GMm}{2R}[/tex]

Therefore, (a) The gravitational potential energy of the system is  [tex]\frac{GmM}{2R^3}r^2 - \frac{3GmM}{2R}[/tex] (Proved), and (b)  the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center will be [tex]-\frac{GMm}{2R}[/tex].

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The complete question is:

a particle of uniform mass m is located inside a uniform solid sphere of radius r and mass m at a distance r from its center (a) sho that the gravitational potential energy of the system is U=GMmr^2 /2R^3 - 3GmM/2R ? (b) Write an expression for the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center.

A rabbit escaping a fox runs across level ground, zipping north for 3.04 m, darting exactly northwest for 2.56 m, and then dropping 0.6 m straight down underground into the safety of its burrow.How far, d, does the rabbit end up from its starting point?

Answers

Zipping 3.04 metres distance in north, precisely 2.56 metres northwest, and finally plunging 0.6 metres straight down into the security of its burrow 4.59 m later, the rabbit has travelled from its starting place.

The rabbit initially travels 3.04 metres north, then 2.56 metres northwest. The Pythagorean theorem may be used to determine the hypotenuse of a right triangle with 3.04 m and 2.56 m-long legs in order to determine the total distance the rabbit has travelled in a north-westerly direction:

d = sqrt(3.04^2 + 2.56^2)

sqrt(15.984), sqrt(9.4304 + 6.5536), and d = 3.99 m

The vertical component of the rabbit's distance travelled must now be included. The rabbit descended 0.6 metres below earth. The rabbit travels a total distance of d = sqrt(15.984) + 0.6 m d = 3.99 m + 0.6 m = 4.59 m from its starting place.

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A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds. How fast was the ball going when it left the boy's hand?

Answers

The ball's exit velocity from the boy's hand was 39.4 m/s.

The time it took the ball to reach its highest point is 4 seconds, which is the explanation.

Use the first equation of motion, v = u + gt, to determine the ball's initial velocity when it left the boy's hand.

Initial velocity (v) is used here. terminal velocity is u. t is the amount of time, and g is the gravitational acceleration, with a value of 9.8 m/s2.

At its highest point, the ball's final velocity, v, equals zero.

When the supplied values are substituted in the equation above, we get 0=u-9.8m/s2.

Therefore, the ball's exit velocity from the boy's hand was 39.4 m/s.

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A proton is fired horizontally into a 1.0×10^5 N/C vertical electric field. It rises 1.0 cm vertically after having traveled 7.0 cm horizontally. What was the proton's initial speed?

Answers

The initial horizontal speed of the proton is 0.16 m/s.

What is the time of motion of the proton?

The time of motion of the proton is calculated by applying the following kinematic equation.

t = √ ( 2gh )

where;

h is the height risen by the protong is the acceleration due to gravity

t = √ ( 2 x 9.8 x 0.01 )

t = 0.44 second

The initial horizontal speed of the proton is calculated as;

d = Vₓt

Vₓ = d / t

Vₓ = ( 0.07 m ) / ( 0.44 s )

Vₓ = 0.16 m/s

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common transparent tape becomes charged when pulled from a dispenser. if one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 13.0 mg piece of tape held 1.30 cm above another. (the magnitude of this charge is consistent with what is typical of static electricity.)

Answers

If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight.

calculate the magnitude of the charge if electrostatic force?

The magnitude of charges on the electron and proton are `1.6xx10^(-19)` C.M Mass of the electrons is `m_(e)=9.1xx10^(-31)` kg and mass of proton is `m_(p)=1.6xx10^(-27)` kg .

n the equation Felect = k • Q1 • Q2 / d2 , the symbol Felect represents the electrostatic force of attraction or repulsion between objects 1 and 2.

The force is perpendicular to both the velocity v of the charge q and the magnetic field B. 2. The magnitude of the force is F = qvB sinθ where θ is the angle < 180 degrees between the velocity and the magnetic field.

This force emerges from the interaction between two charged objects (or point charges) and its magnitude is calculated by F=kQ1Q2r2 F = k Q 1 Q 2 r.

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which of the following is the best way driver/operators can prevent the aerial device from moving or chattering along a frozen solid surface on an angle? group of answer choices using wheel chocks deploying manual stabilizers positioning the apparatus laterally positioning the apparatus longitudinally

Answers

Using wheel clocks is the best option.

To stop an aerial device from moving or chattering on an angled frozen solid surface, wheel chocks are intended to be installed in front of and behind the wheels of the aerial device. The aerial equipment can't move or chatter since the chocks mechanically stop the wheels from rotating. Additionally, they can be utilised to keep the apparatus stable on the surface by adding an additional layer of friction between the ground and the object. Wheel chocks are an efficient way to stop aerial equipment from shifting or chattering on an angled frozen solid surface when used in conjunction with hand stabilisers.

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How far (in m) does a car going 21 m/s travel in 7.1 s?

Answers

Use this formula

S = ut + (1/2)at²

Where, S = Diatance. u = Initial Velocity. a = acceleration.

There is no acceleration.So we can use S=ut formula directly.

S=ut
= 21 x 7.1
= 149.1

The answer is 149.1 m

A body of mass 50 kg explodes and splits into three pieces. The first piece has a mass of 10 kg and a velocity of [-3,2] m/s, the second piece has a mass of 18 kg and a velocity of [5, -4] m/s. What is the velocity of the third piece?
. ​

Answers

The velocity of the third piece 2/11. (13 j - 15i)

What is velocity?Velocity is the directional velocity of a moving object  as an indicator of the rate of change of position  observed from a particular frame of reference and  measured by a particular time standard.Velocity is a vector representation of the displacement  an object or particle experiences with respect to time. The standard unit for velocity magnitude (also called velocity) is meters per second (m/s). Alternatively, centimeters per second (cm/s) can be used to express velocity magnitude.Simply put, velocity is the speed at which something moves in a particular direction. For example, the speed of a car traveling north on a highway, or the speed of a rocket  after launch.

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Forces at Work
The spring-loaded cart is a simple system, which can make it easier to
analyze and apply the scientific ideas observed to a larger, more
complex system. A skateboarder is another simple system. A
skateboarder would not get anywhere without forces. Since the
skateboard does not have a motor, the skateboarder must be the one
that supplies the power to make it go. But how does this push cause
the motion? Why does the skateboarder have to continually push the
skateboard to keep it in motion?
Similar cause-and-effect relationships involved in the spring-loaded
cart can be applied to your model of the fire extinguisher go-kart.
What cause-and-effect relationships identified in
the spring-loaded cart can be applied to the
rocket sled?
Effect
Cause
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Cause
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Effect
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Answers

The skateboard was initially at rest, and by the time of Newton's first law, it would still be in that state. This law asserts that unless acted upon by an external or net force, a body will remain at rest or continue in a condition of uniform motion with constant velocity (in a situation where not only one force is in action).

According to this law, in order to move the skateboard, it needs an external source of force, which the skateboarder indirectly provides.

The skateboarder reverses forward on the pavement (that is he applies a force on the road in a direction opposite the direction of intended motion).

According to Newton's third law, the skateboarder's action results in an equal and oppositely directed reaction from the road on the skateboarder.

According to Newton's third law, action and response are equal but directed in opposite directions. As a result, the road pulls the skateboarder forward in response to this rearward force, which moves both the skateboarder and the skateboard in the same direction.

The acceleration of the cart is inversely related to its mass. As a result, the acceleration of the cart will decrease as its mass increases.

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A barrel rider is moving in a circle with a speed of 7.0 m/s. The acceleration of the rider is 7.2 m/s/s. The speed of the object is somehow increased to 14.0 m/s (i.e., doubled). The new acceleration would be _____ m/s/s. (Assume that the radius of the circle is not changed.)

Answers

The new acceleration of the barrel rider if the speed is doubled is 14.4m/s².

How to calculate acceleration?

Acceleration in physics refers to the amount by which a speed or velocity increases i.e. the change of velocity with respect to time.

The acceleration of a moving body is directly proportional to the speed of that body i.e. an increase in speed equates to an increase in acceleration.

According to this question, a barrel rider is moving in a circle with a speed of 7.0 m/s. The acceleration of the rider is 7.2 m/s². However, the speed of the object is somehow increased to 14.0 m/s.

This suggests that the acceleration of the rider will also be doubled and hence, be 14.4m/s².

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A certain amount of heat is added to some water so that its temperature rises. The same amount of heat is added to a piece of alu- minum with the same mass as the water. Which has the higher temperature change? A)water B) aluminum C) they gave equal temperature changes​

Answers

At upper temperatures, the thermal conductivity decreases slowly with increasing temperature than depicted by this equation. Aluminum  has the higher temperature change.

What is Aluminum  ?Aluminum is a silvery-white metal and is the 13th element on the periodic table. A surprising fact about aluminum is that it is the most common metal on earth.Aluminum (Al), also known as aluminum, chemical element, light silvery-white metal of the 13th main group (III a, or boron group) of the periodic table. Aluminum is the most abundant metallic element in the earth's crust and the most commonly used non-ferrous metal. Aluminum is a silver-white colored lightest metal. Soft and easy to play. Aluminum is used in a  variety of products including cans, foils, kitchenware, window frames, beer kegs and aircraft components.

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according to a newspaper account, a paratrooper survived a training jump from 1156 ft when his parachute failed to open but provided some resistance by flapping in the wind. allegedly he hit the ground at 99.25 mi/h after falling for 10 seconds. to test the accuracy of this account, you should first find the drag coefficient , assuming a terminal velocity of 99.25 mi/h and also that the deceleration of the paratrooper due to air resistence is proportional to his velocity, with constant of proportionality . remember that the acceleration due to gravity near the earth's surface is 32 ft/sec
a. Find rho.
b. Next, find the distance fallen in 6 seconds..

Answers

An object can encounter some resistance when moving through the fluid medium of air.

The drag force is the oppositional force applied by the air to a moving object. The size of the cross section A of the front face of the item, the density of the air, and the square of the object's velocity v all contribute to the drag force.

With motion exclusively in the vertical direction and letting

downward be positive, we have that

dv/dt = g - pv, where g = 32 ft/[tex]s^{2}[/tex], and so

dv/dt + pv = g. The integrating factor is [tex]e^{(pt)}[/tex], so

d/dt([tex]e^{(pt)}[/tex] * v) = g×[tex]e^{(pt)}[/tex] ----->

[tex]e^{(pt)}[/tex] * v = (g/p)*[tex]e^{(pt)}[/tex] + C ----->

v(t) = (g/p) + C×[tex]e^{(pt)}[/tex]

Now v(0) = (g/p) + C = 0, and so C = -g/p, giving us

v(t) = (g/p)×(1 - [tex]e^{(pt)}[/tex]).

Now we are given that lim(t->infinity)(v(t)) = 99 mi/h = 145.2 ft/s,

so since [tex]e^{(pt)}[/tex]-> 0 as t -> infinity we have g/p = 145.2 ft/s ------>

p = 32 ft/s^2 / (145.2 ft/s) = 0.2204 s^-1 .

Next, since we have defined downward as positive, the expression

we have for v(t) will equal dy/dt where y(0) = 0 and y increases as

the paratrooper falls. So

dy/dt = v(t) = (g/p)×(1 - [tex]e^{(pt)}[/tex])), so

y(t) = (g/p)×(t + (1/p)×[tex]e^{(pt)}[/tex]) + K.

Now y(0) = (g/p)×(0 + (1/p)) + K = 0 ----->

K = -(g/p^2) = 32 / 0.2204^2 = -658.76 ft, so

y(t) = (145.2)×(t + (4.5372)×e^(-0.2204×t)) - 658.76.

Next, y(6) =  (145.2)×(6 + (4.5372)×e^(-0.2204×6)) - 658.76 = 444.84 feet.

The computed drag coefficient is substantially higher than anticipated.

The free fall distance is the longest distance an individual can travel in a predetermined amount of time. However, the distance covered in the time frame indicated in the news report is significantly greater than the greatest distance that could be covered. Consequently, the news story is exaggerated.

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Which of the following will have the highest surface tension?
A
Water
B
Ethanol
C
Methanol
D
Ethanal

Answers

Water has the highest surface tension than ethanol, methanol, or ethanal.

The cohesive nature of a liquid's molecules gives rise to a property of the liquid's surface known as surface tension, which enables the liquid to resist the action of an outside force. After mercury, water has the highest surface tension of any liquid. This is because hydrogen bonds are present in water molecules, which gives water its high surface tension. As a result of water's surface tension, water molecules that are in close proximity to one another at the liquid's surface (where they are in contact with air) stick together to form an invisible film. When water is heated to 25 degrees Celsius, its surface tension is 72 dynes/cm. To crack a film of water on the surface that is one centimeter long and requires 72 dynes of force.

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Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 5.5 m/s to 10.4 m/s.

Answers

The stone will descend straight down for 0.5 seconds, increasing its speed from 5.5 m/s to 10.4 m/s.

Newton's second law of motion states that F = ma where ma is the object's mass and an is its acceleration.

You can also write the equation as:

F = m (v - u)/t

Ft =m (v-u)

mat = m (v-u)

at = v-u

t = v-u/a

Given the following parameters;

v = 10.4 m/s

u = 5.5 m/s

a = 9.8m/s²

Substitute the given parameters into the formula to have:

t = 10.4-5.5/9.8

t = 4.9/9.8

t = 0.5 sec

The stone's speed will therefore increase from 5.5 m/s to 10.4 m/s in 0.5 seconds as it drops directly to the ground.

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