state two factors that affect current carrying capacity of an accumulator

Answers

Answer 1

The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

What is the function of the accumulator?

Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.

These key deciding elements are:

1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.

2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.

3. Accumulator identification:

4. Conditions for Installation:

Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

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Related Questions

the road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60 km / h. Find the average round trip speed.​

Answers

Answer:

Let distance between two cities is x km . So total distance travelled bus x+x=2x km and time taken was=(x/40)+(x/60)=(3x+2x)/120=5x/120=x/24 hrs . So avg speed is 2x/(x/24)=48 km per hr.

or

dt1=60km1h

dt2=30km1h

2dt1+t2=r¯

d=60t1

d=30t2

60t1=30t2

2t1=t2

2dt1+(2t1)=r¯

2(60t1)t1+2t1=r¯

2(60t1)=r¯(t1+2t1)

120=r¯+2r¯

r¯=1203

r¯=40km/h

Identify the law, write the equation and calculate the answer to the problem below.

An incident light ray strikes water at an angle of 45 degrees. The index of refraction of air is 1.0003, and the index of refraction of water is 1.33. What is the angle of refraction rounded to the nearest whole number?

Answers

Find refractive index first

[tex]\\ \rm\Rrightarrow \mu=\dfrac{c}{v}[/tex]

[tex]\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}[/tex]

[tex]\\ \rm\Rrightarrow \mu =0.75[/tex]

Now

[tex]\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr[/tex]

[tex]\\ \rm\Rrightarrow sinr=0.94[/tex]

[tex]\\ \rm\Rrightarrow r=sin^{-1}(0.94)[/tex]

[tex]\\ \rm\Rrightarrow r=70^{\circ}[/tex]

Charlene has just begun to be able to form a mental representation of an object that is not visibly
present According to Piaget, that means that she has made the transition from the
stage
to the
stage

Answers

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

What is the sensorimotor stage?

The first of Piaget's four stages of cognitive development is the sensorimotor stage.

A child's understanding that the outside world exists apart from them is what distinguishes it.

Within Piaget's stages of development, the kid will advance to the following stage after they have completely grasped this.

"Charlene has just begun to be able to form a mental representation of an object that is not visibly present.

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

Only recently has Charlene been able to create a mental image of something that is not physically present.

That indicates that she has moved from the sensorimotor to the preoperational level, in accordance with Piaget.

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A mass of (200 of hot water at (75, 0 degrees * C) is mixed with cold water of mass M at (5, 0 degrees * C) The final temperature of the mixture is (25, 0 degrees * C) What is the mass of the cold water (M)

Answers

Q in = Q out

mcΔt in = mcΔt out (c = equal)

mΔt in = mΔt out

200 x  (75-25) = M x (25-5)

M = 500 g

∆T for hot water

75-2550°C

∆T for cold water

25-520°C

Both are equal according law of conservation of energy

[tex]\\ \rm\Rrightarrow m_1c\delta T_0=m_2c\delta [/tex]

[tex]\\ \rm\Rrightarrow 200(50)=m_2(20)[/tex]

[tex]\\ \rm\Rrightarrow 10000=m_2(20)[/tex]

[tex]\\ \rm\Rrightarrow m_2=500g=0.5kg[/tex]

(the problem setup and question are below)

Answers

The gauge pressure at bottom of vaccine solution will be 16 kPa

Positive pressure is another name for gauge pressure. When a system's internal pressure exceeds that of its surroundings, it is said to be under positive pressure. Any leak that develops in the positively pressured system will therefore escape into the outside world. In contrast, a negative pressure chamber draws air into it.

Given As seen in the illustration, a syringe is held vertically. The container carries a 3 cm tall column of vaccine solution and has an open inner diameter of 1 cm. The needle contains a 2 cm column of vaccine solution and has an open inner diameter of 0.5 mm. At the needle's open end, the solution is exposed to the air. The vaccination solution has a density of 1200 kg/m3.

We have to find the gauge pressure at bottom of vaccine solution

Since the 5N force is applied to vaccine solution the pressure exerted will be much more

Hence the gauge pressure at bottom of vaccine solution will be 16 kPa

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A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 31 N. Starting from rest, the sled attains a speed of 2.3 m/s in 9.2 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

What is the number and units?

Answers

The sled accelerates with magnitude [tex]a[/tex] such that

[tex]\left(2.3\dfrac{\rm m}{\rm s}\right)^2 = 2a(9.2\,\mathrm m) \implies a = 0.2875\dfrac{\rm m}{\mathrm s^2}[/tex]

By Newton's second law, the net force in the plane of motion (parallel to the ground) is

[tex]31\,\mathrm N - F_{\rm friction} = (17 \,\mathrm{kg}) \left(0.2875 \dfrac{\rm m}{\mathrm s^2}\right)[/tex]

so that the force of friction exerts a magnitude of

[tex]F_{\rm friction} = 26.1125 \,\mathrm N[/tex]

Perpendicular to the ground, the sled is in equilibrium, so Newton's second law says

[tex]F_{\rm normal} - (17\,\mathrm{kg})g = 0 \implies F_{\rm normal} = 166.6 \,\mathrm N[/tex]

The magnitude of friction is proportional to the magnitude of the normal force by a factor of [tex]\mu_k[/tex], the coefficient of kinetic friction. It follows that

[tex]F_{\rm friction} = \mu_k F_{\rm normal} \implies \mu_k = \dfrac{26.1125\,\rm N}{166.6\,\rm N} \approx \boxed{0.16}[/tex]

(and the coefficient is dimensionless).

The action force is the ballon pushing the air out. What is the magnitude of the reaction force of the air pushing the ballon

Answers

The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.

How is the reaction force equal and opposite?

A push or a pull that an object experiences as a result of interacting with another item is known as a  reaction force.

Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.

Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.

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A solenoid with 465 turns has a length of 8.00 cm and a cross-sectional area of 3.10 ✕ 10−9 m2. Find the solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s

Answers

The solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s will be 1.01 ×10⁻⁸ H and 9.02 × 10⁻² V.

What is a solenoid?

A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire.

Which generates a magnetic field when an electric current is passed through the coil.

A solenoid is a form of coil that produces a magnetic field when the electric current is passed through it. A solenoid is created when a conductive wire is used to make a loop.

Given data;

Turns ,N = 465

Length,L= 8.00 cm

Cross-sectional area,A = 3.10 ✕ 10−9 m2.

Solenoid's inductance, L=?

The average emf around the solenoid, E=?

Time of flow,t= 7.83 ✕ 10−3 s

Current changes from +3.50 A to −3.50 A

The inductance of the solenoid is found as;

[tex]\rm L = \frac{\mu_0 AN^2}{L} \\\\ \rm L = \frac{4 \times \pi \times 10^{-7}\times 3.0 \times 10^{-9} \times (465)^2 }{8.00 \times 10^{-2}} \\\\ L= 1.01 \times 10^{-8} \ H[/tex]

The average emf around the solenoid is found as;

[tex]\rm e = L \frac{I_2-I_1}{t} \\\\ \rm e = 1.01 \times 10^{-8} \times \frac{3.50-(-3.50)}{7.83 \times 10^{-3}} \\\\ e =9.02 \times 10^{-12} \ V[/tex]

Hence, the solenoid's inductance will be 1.01× 10⁻⁸ H.

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a generator produces a maximum emf of 240 v when rotating at 3.7 rad/s in a magnetic field of 0.200 t. if the coil has 822 loops what is the area of each loop?

Answers

Answer:

I actually did that but I fprgot

Answer: 0.0387

Explanation: I had the same problem and here's what I got:

Values:

ε (Voltage) = 240

ω (Angular velocity) = 37.7

B (Magnitude) = 0.200

N (Number of loops) = 822

ε = NABω (Rearrange to get A)

A = ε/WBN

= (240)/(37.7)(0.200)(822)

A = 0.0387 (FINAL ANSWER)

A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C

Answers

The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

What is the specific heat capacity?

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol  m₁ = 200-gram

The temperature of alcohol, T₁ =  -6°C.

Mass of liquid sample of water  m₂ = 400-gram

The temperature of the water, T₂=  20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]

Hence the specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

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A block of mass 4 kg is attached to a spring of force constant (k) 1000 N/m, and undergoes simple harmonic motion. What is the period of the motion?
a. 0.1 s
b. 0.4 s
с. 0.2 s
d. 0.3 s​

Answers

Answer:

0.4

Explanation:

ω = km

ω = 1000 x 4

ω = 4000 divide this by 10000 and you get 0.4s

Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c. C d. D

Answers

Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

Which graph shows the correct relationship between kinetic energy and speed? A. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A straight line starts at the y-axis B. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A straight line starts from the origin C. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A curve line runs parallel to the x-axis and y-axis D. Graph representing a relationship between speed on the x-axis and kinetic energy on the y-axis. A semi-curve line starts above the origin on the y-axis and curves upwards as it moves forward in speed

Answers

The proper connection between kinetic energy and speed is depicted in Graph D.

What is kinetic energy?

The kinetic energy (KE) is defined as one-half of the mass times multiplied by the square of velocity.

[tex]\rm KE = \frac{1}{2}mv^2[/tex]

where,

KE is the kinetic energy

m is the mass of each molecule

(V) is the  velocity

As the square of the velocity is exactly proportional to kinetic energy. Consequently, the relationship between velocity and kinetic energy must be parabolic.

The proper connection between kinetic energy and speed is depicted in Graph D. Graph showing the link between kinetic energy on the y-axis and speed on the x-axis.

On the y-axis, a semi-curve line begins above the origin and ascends as it accelerates.

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'The wave' is a particular type of pulse that can propagate through a large crowd gathered at a sports arena to watch a soccer, hockey, or CFL game. The elemets of the medium are the spectators, with zero position corresponding to their being seated and maximum position corresponding to their standing and raising their arms. When a large fraction of the spectators participate in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people's reaction time, which is typically on the order of 0,1s. Estimate the order of magnitude, in minutes, of the time required for such a pulse to make one circuit around BC Place Stadium in Vancouver.

State all the assumptions that you've made.

Information about BC Place: dimensions are approximately 100 m X 85 m.

Answers

The total time required is 1 minute and 0.3 seconds.

Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.

Here, the time taken by the person is 0.1 seconds.

Total distance covered by the wave = Circumference of the circle

So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m

As the distance between each of the people is 1 m.

So, the number of personal interactions is 618.

Time taken by each person is 0.1 seconds.

So, total time, t = 0.1 × 618 = 61.8 seconds.

So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.

Hence, the total time required is 1 minute and 0.3 seconds.

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Which of the following is not a use of nuclear chemistry in industry?

thickness gauges
tracers
leak detectors
stimulate plant growth

Answers

Stimulating plant growth is not a use of nuclear chemistry in industry. Thus, the correct option is D.

What is Nuclear Chemistry?

Nuclear chemistry may be defined as the sub-branch of chemistry that involves the investigation of physical and chemical characteristics of elements that are affected by transformations in the configuration of the atomic nucleus.

Thickness gauges measure the thickness, density, and fill levels and are utilized in manufacturing to make sure that an entire product or a set of products are of identical thickness.

Tracers are utilized in tracing the actual composition of radioactive decay through the chemical compounds. While the leak detectors monitor the actual and constant flow of water through a given system precisely.

Therefore, the correct option for this question is D.

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Answer:

it is d

Explanation:

Determine the amount of power
used in holding a 25 kg box, 1.5
meters above the floor, for 60
seconds.

[?] W

(answer is not 6.13)

Thank you in advance!

Answers

Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

Power is equal to work done per unit time

Work is force × displacement

SI UNIT OF WORK Newton meter

SI UNIT OF POWER Watt

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot​

Answers

The force the rod exerts on the ball at the given angle is determined as 3.94 N.

Force exerted on the rod by the ball

The force exerted is calculated as follows;

F = ma

F = mv²/r

F = mω²r

where;

m is mass of the ballω is angular speed of the ballr is radius of the path

r = 2cosθ

Angular speed when the ball moves 15 degrees

ωf² = ωi² + 2αθ

where;

θ is angular displacement in radians, 15⁰ = 15 x π/180 rad

ωf² = (1)² + 2(2)(15 x π/180)

ωf² = 2.04

ωf = √2.04

ωf = 1.428 rad/s

F = mω²(2cosθ)

F = (1)(1.428)²(2 x cos15)

F = 3.94 N

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You drive 8.50 km in a straight line in a direction 30° East of North.
(a) Find the distances you would have to drive straight East and then straight North to arrive at the same point. (This is equivalent to finding the components of the displacement along the East and North directions.)
km East
km North
(b) Show that you still arrive at the same point if the East and North legs are reversed in order.

Answers

(a.1) The component of displacement along the East is  7.36 km.

(a.2) The component of displacement along the North is  4.25 km.

(b)  When the East and North legs are reversed in order, you will still travel 8.5 km.

Component of displacement along the East

dx = d cosθ

dx = 8.5 km x cos(30)

dx = 7.36 km

Component of displacement along the North

dy = d sinθ

dy = 8.5 km x sin(30)

dy = 4.25 km

When East and North legs are reversed;

dx = 4.25 km

dy = 7.36 km

Resultant displacement;

R = √(dx² + dy²)

R = √(4.25² + 7.36²)

R = 8.5 km

Thus, when the East and North legs are reversed in order, you will still travel 8.5 km.

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Which picture shows the wave with the smallest wavelength?
Select one:
a. A
b. B
c. C
d. D

Answers

Answer:

b. B

Explanation:

Picture B has the smallest peaks among all which henceforth makes the wavelength i.e. distance between two adjacent crests or troughs, small.

Answer:b

Explanation:Wavelength can be calculated using the following formula: wavelength = wave velocity/frequency. Wavelength usually is expressed in units of meters. The symbol for wavelength is the Greek lambda λ, so λ = v/f.

A contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake as shown in Figure 4.29(a). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
a) Calculate the minimum force F he must exert to get the block moving.
N
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2

Answers

Answer:

54.0 x0.1=5.4 x0.03=0.162

kinetic force

If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.

What is friction?

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,

The force required to get the block moving = μMg

                                                                          = 0.01×54×9.81

                                                                         = 5.2974 Newtons

Thus, the minimum force required to move the block would be 5.2974 Newtons.

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the primary coil of a transformer has 480 turns, and the secondary has 112 turns. when an ac voltage of 28.0 v is applied to the primary, what is the output voltage in the secondary?

Answers

The voltage of the secondary will be 36 V.From the given conditions primary transformer has 3 times as many turns in the secondary coil.

What is induced voltage?

Electromagnetic induction is what causes the induced voltage. Electromagnetic induction is the process of generating emf (induced voltage) by subjecting a conductor to a magnetic field.

Given data;

No turns in the primary coil,[tex]\rm N_p = 480 turns[/tex]

No turns in the secondary coil,[tex]\rm N_s = 112 turns[/tex]

The voltage of the primary coil,[tex]\rm V_p = 28 v[/tex]

The voltage of the secondary coil,[tex]\rm V_s = ?[/tex]

For a transformer,

[tex]\rm \frac{V_p}{V_s}= \frac{N_p}{N_s} \\\\ \rm \frac{28}{V_s}= \frac{480}{112} \\\\ V_s = 6.53 \ v[/tex]

Hence the voltage of the secondary will be 36 V.

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Answer: 6.53

Explanation: I had the same question and here's what I got:

Values:

Vp (Primary Voltage) = 28.0

Ns (Secondary Number of Turns) = 112

Np (Primary Number of Turns) = 480

Vs = Vp Ns/Np

= (28)(112)/(480)

Vs = 6.53 (FINAL ANSWER)

When passing in a medium for a distance of 1.5 cm intensity

of the light decreased by 3 times. What will the distance x equal to when the intensity of the light decreases by 9 times?

Answers

The distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

Intensity of light

The intensity of light is given as power emitted by the light by unit area.

I = P/A

I = P/L²

I₁L₁² = I₂L₂²

L₂² = I₁L₁²/I₂

L₂² = (3 x 1.5²)/(9)

L₂² = 0.75

L₂ = 0.87 cm

Thus, the distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

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Your heart pumps about 70 mL of blood per beat and beats an average of 80 Times per minute how many gallons of blood does your heart pump every hour

Answers

Answer:

88.761

Explanation:

What is the acceleration of the object?
m/s²

Answers

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

Question 4 of 20
If the distance between two objects is doubled, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It increases to 2 times its original magnitude.
C. It increases to 4 times its original magnitude.
D. It decreases to 1/4 its original magnitude.

Answers

Answer:

D

Explanation:

F = G m1 m2 / r^2       now double  r

 F =  G m1m1/ (2r)^2

    F =  1/4  G m1m2/r^2     <===== this is 1/4 of the original

A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board. He walks from one end of the board to the other end with a velocity of 1.47 m/s relative to the ice in the positive direction. What is the velocity of the board relative to the ice?

Answers

The velocity of the board relative to the ice is zero, since both are at rest.

What is relative velocity?

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

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A skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally.
H = 1.15x10^1 m, h= 6.50 m, and D = 11.4 m.
Find her total kinetic energy as she reaches the ground.

Answers

The total kinetic energy of the woman when she reaches the ground is 6,198.5 J.

Total kinetic energy of the woman at the ground level

The total kinetic energy of the woman when she reaches the ground is calculated from the principle of conservation of energy.

P.E(top) = K.E(bottom)

P.E(top) = mgh

P.E(top) = 55 x 9.8 x 11.5

P.E(top) = 6,198.5 J

Thus, the total kinetic energy of the woman when she reaches the ground is 6,198.5 J.

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R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the details of the equation; the basic point is that if you put an amount of mass M or larger in a sphere of radius smaller than RS(M)
What equation is this

Answers

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

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Question 2 of 25
What is the term for a pair of forces described by Newton's third law?

A.
Normal-friction pair

B.
Action-reaction pair

C.
Kinetic-static pair

D.
Inertia-movement pair

Answers

The term for a pair of forces described by Newton's third law is action-reaction pair. Details about Newton's law can be found below.

What does Newton's third law state?

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

This law proposed that when an object 1 acts on object 2 with a force of a particular magnitude, object 2 also acts on object 1 with an opposite force of same magnitude.

Therefore, it can be said that the term for a pair of forces described by Newton's third law is action-reaction pair.

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4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The
horizontal range of the ball is R, and the ball reaches a maximum height R
6
. In
terms of R and g, find (a) the time interval during which the ball is in motion,
(b) the ball’s speed at the peak of its path, (c) the initial vertical component of
its velocity, (d) its initial speed, and (e) the angle θi

Answers

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by

[tex]x = v_i \cos(\theta_i) t[/tex]

[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]

and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are

[tex]v_x = v_i \cos(\theta_i)[/tex]

[tex]v_y = v_i \sin(\theta_i) - gt[/tex]

The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when

[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]

which means

[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]

At the same time, the ball will have traveled half its horizontal range, so

[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]

Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :

[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]

Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)

[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]

Now,

[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]

so it follows that (d)

[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]

Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is

[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]

Finally, in the work leading up to part (e), we showed the time to maximum height is

[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]

but this is just half the total time the ball spends in the air. The total airtime is then

[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]

and the ball is in the air over the interval (a)

[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]

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