2. A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0° above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

Answers

Answer 1

Answer: See below

Explanation:

Part (a):

As the velocity of the piano is constant, the net force on the piano is zero. The friction is also zero.

Here, F is the force applied by the man towards the inclined plane, mg is the weight of the piano and N is the normal force.

Applying Newton's law we get,

[tex]F = mg\sin \theta[/tex]

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \sin 19.0^\circ \\ &= 574.3\;{\rm{N}}[/tex]

Therefore, the force is 574.3 N

Part (b)

Here, the force F is applied parallel to the floor.

The friction is zero.

Applying Newton's law we get,

[tex]F\cos \theta &= mg\sin \theta \\ F &= mg\tan \theta[/tex]

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \tan 19.0^\circ \\ &= 607.4\;{\rm{N}}[/tex]

Therefore, the force is 607.4 N


Related Questions

A penny is dropped so that it reaches a terminal velocity of 13.3m/s. If the momentum is measured to be 0.0332kgm/s, what is the mass of the penny

Answers

Answer:

2.5 grams

Explanation:

p = mv

m = p/v

m = 0.0332 / 13.3

m = 0.0025kg

0.0025 x 1000 = 2.5

m = 2.5g

suppose the velocity of a particle is given by the equation, V=m+nt2, where m=10cms and n=2cm/s3.
A, find the change in velocity of the particle in the initial time interval b/n t₁=2sec and t₂=5sec
B, find the average acceleration in this time interval
C, find the instantaneous acceleration at time t₁=2sec​

Answers

Hello!

Begin by plugging in the values for m and n. We get the equation for the velocity of the particle to be:
[tex]v(t) = 10 + 2t^2[/tex]

A.

To find the change in velocity over the interval (2s ≤ t ≤ 5s), we can simply find the difference in the velocities at these times.

[tex]\Delta v= v_f - v_i[/tex]

For this situation:
[tex]\Delta v = v(5) - v(2)[/tex]

Substitute these times for 't' into the equation and solve.

[tex]v(5) = 10 + 2(5^2) = 60 \frac{cm}{s}\\\\v(2) = 10 + 2(2^2) = 18 \frac{cm}{s}\\\\\Delta v = 60 - 18 = \boxed{42 \frac{cm}{s}}[/tex]

B.

To find the average acceleration, we must take the SLOPE of the velocity function over this interval using the slope formula:

[tex]a_{avg} = \frac{v_f - v_i}{\Delta t}[/tex]

Plug in the values for the particle's velocity at t = 2 s and 5 s that we solved for above.

[tex]a_{avg} = \frac{60- 18}{5 - 2}\\\\a_{avg} = \frac{42}{3} = \boxed{ 14 \frac{cm}{s^2}}[/tex]

C.

The instantaneous acceleration can be found by taking the derivative of the v(t) function using the power rule. Recall:
[tex]\frac{dy}{dx} x^n = nx^{n-1}[/tex]

Using this rule:
[tex]a(t) = v'(t) = 2(2t) = 4t[/tex]

Substituting in t = 2 s:
[tex]a(2) = 4(2) = \boxed{8 \frac{cm}{s^2}}[/tex]

A mass of 2.0 kg of water is heated. The temperature increase of the water is 80 degrees Celsius. The specific heat capacity of water is 4200 J / kg degrees Celsius. Calculate the change in thermal energy when the water is heated. Use the equation. change in the thermal energy = mass x specific heat capacity x temperature change

Answers

Answer:

672 000 J      or 672 kJ

Explanation:

change in the thermal energy = mass x specific heat capacity x temperature change

     =   2 kg * 4200J/kg-C *80 C =672000 J

Question 3 of 10
The Moon has much less gravitational force than Earth. What would happen
if you went to the Moon?
OA. Your mass would decrease.
O
B. Your weight would increase.
O
C. Your mass would increase.
O
D. Your weight would decrease.
SUBMIT

Answers

Answer:

D

Explanation:

Mass is the amount of matter your body has ....does not change on the Moon.

there is less gravity .....so your WEIGHT would decrease

which of the following statements is true about the suns energy and earth​

Answers

The correct statement about the sun and the earth is that only about half of the suns energy hits the earths surface.

What is the sun?

The sun is the source through which energy reaches the earth. The sun is a large star that produces energy by the process of nuclear fusion.

Hence, the correct statement about the sun and the earth is that only about half of the suns energy hits the earths surface.

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Compute the speed of sound waves in air at room temperature (T=200C) and find the range of wavelengths in air to which the human ear (which can hear frequencies in the range of 20 – 20,000Hz) is sensitive. The mean molar mass of air (a mixture of principally nitrogen and oxygen) is 28.8 x 10-3kg/mol and the ratio of heat capacity is ϒ = 1.40.

Answers

The speed of sound waves in air at room temperature T=20° is 344.035 m/s.

The range of wavelength in air to which the human ear (which can hear frequencies in the range of 20 – 20,000Hz) is sensitive is  17.20 m to  0.0172 m.

What is wavelength?

The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.

Wavelength λ = c/f

where c is the speed of sound wave.

Given the room temperature T = 20 +273 = 293K

The velocity of sound wave is given by

v = √(γRT/m)

​where γ =1.4 , m = 28.8×10⁻³ kg/mol  R = 8.31 J/K.mol

Substituting the values, we get

v = 344.035 m/s

Thus, the speed of sound wave is 344.035 m/s

The human ear can hear frequencies in the range of 20 – 20,000Hz.

The wavelength corresponding to 20Hz is

λ = c/f

λ =  344.035 /20

λ  = 17.20 m

The wavelength corresponding to 20,000Hz is

λ =  344.035 /20000

λ  = 0.0172 m

Thus, the range of wavelength is 17.20 m to  0.0172 m sensitive to human ear.

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The three waves shown in the picture below are equal in wavelength, but are travelling at different speeds. Given this information, which wave has the highest frequency?
Select one:
a. 1
b. 2
c. 3
d. Can't tell from the information given.

Answers

Wave 3 will have the highest frequency. Option c is correct.

What is the frequency?

Frequency is defined as the number of repititions of a wave occurring waves in 1 second.

Frequency is given by the formula as,

[tex]\rm v = \lambda \times f[/tex]

If the wavelength of the wave is content the velocity of the wave is directly propotional to the frequency.

The velocity of wave 3 is maximum. So that the wave 3 will have the highest frequency.

Hence option c is correct.

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pleaseeeeeee I need HELP in 20 mins,Asap!!!


A 6.00 -kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22m/s^2 and shatters into three pieces, which all fly backward. The wall exerts a force of 2640N on the ball of for 0.1s. One piece of mass 2kg travels backward at a velocity of 10m/s and an angle of 32° above the horizontal. A second piece of mass 1kg travels at a velocity of 8m/s and an angle of 28° below the horizontal. What is the velocity of the third piece?​

Answers

The velocity of the third piece is 124.02 m/s at 1.05⁰ below the horizontal.

Velocity of the third piece

The velocity of the third piece is calculated from the principle of conservation of linear momentum.

mu = m₁u₁ + m₂u₂ + m₃u₃

where;

m is mass of the clayu is velocity of the clayu₁ is velocity of first pieceu₂ is velocity of second pieceu₃ is velocity of third piecem₃ is mass of the third piece = 6 kg - (2 kg + 1 kg) = 3 kgMomentum in y - direction

6(22)sin(0) = 2(10)sin32 - 1(8)sin(28) + 3u₃y

0 = 6.84 + 3u₃y

u₃y = -6.84/3

u₃y = -2.28 m/s

Change in momentum

ΔP = Pf - Pi = J

where;

Pf is final momentumPi is the initial momentumJ is impulse

2640(0.1) = 2(10)cos32 +  1(8)cos(28) + 3u₃x - 6(22)

264 = -108 + 3u₃x

3u₃x = 372

u₃x = 372/3

u₃x = 124 m/s

Resultant velocity

u₃ = √(124² + 2.28²)

u₃ = 124.02 m/s

Direction of the velocity

tanθ = u₃y/u₃x

tanθ = 2.28/124

tanθ = 0.018

θ = 1.05⁰ (below the horizontal)

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What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the ground?
-441J
735J
-735J
441J

Answers

Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J


On Earth, a scale shows that you weigh 490 N. What is your mass?
A 50 kg
B 100 kg
c) 75 kg
D 22 kg

Answers

A 50 kg

490N = m(kg) *9.8(m/s^2)=>m=490/9.8 = 50kg

Answer:

a

Explanation:

A pendulum swings to and fro every 2.0 s. The period of the swing is.
O 2 Hz
2 s
O 0.5 s
O 0.5 Hz

Answers

Answer:

Explanation:

Solution

In physics, a period is a measurement of time. The answer must be either B or C. It is a direct measurement of time, not its reciprocal. So the answer is B. A and D is a measurement of the actual cycle that it takes before the motion begins to repeat itself.

Answer

B

Helium is the second most abundant element on Jupiter (and in the Universe overall). But it's rare on Earth, being only 0.0005% of the atmosphere. Why is this?

A) Jupiter's stronger gravity results in a faster escape velocity, and almost no helium atoms are moving fast enough to escape its gravity. Earth's escape velocity is lower, so a few helium atoms at a time are moving fast enough to escape, causing Earth to slowly lose its helium.
B) Jupiter's core continually generates helium, while the Earth's core does not.
C) Jupiter has a powerful magnetic field that can attract helium atoms. Earth has a magnetic field as well, but it's not strong enough.

Answers

Answer:

Explanation:

Comment

I think the best answer is probably A.  It's not the best reason for Earth loosing it's helium, but it is the only one that is close. The Earth derives some of its Helium (most) by the decomposition process of radio active high weight chemical that break down. Helium combines with practically nothing so once it gets moving, very little will stop it. That's the way we loose our Helium.

Still answer A.

an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of the object?

Answers

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

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How to magnetic stripes provide evidence of seafloor spreading

Answers

Magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

How to magnetic stripes provide evidence of seafloor spreading?

When the Earth's magnetic field reverses, a new stripe starts to form. Such magnetic patterns enable the scientist to recognize the occurrence of sea-floor spreading and provides strongest evidence for the theory of plate tectonics.

So we can conclude that magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

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(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?​

Answers

(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]

where [tex]R[/tex] is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]

so that

[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]

Solve for [tex]v[/tex] :

[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]

(b) The net force equation in part (a) leads us to the relation

[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]

so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.

A radioactive contaminant gives an unfortunate lab rat a dose of 1500 rem in just 1 minute. Assuming that the half life of the radioactive isotope in the contaminant is much longer than 1 minute.What would the activity of the contaminant be if the contaminant is 1.1MeV Beta Emitter?

Answers

The activity of the contaminant will be 3.5 ×10⁻¹⁰ Bq if the contaminant is 1.1MeV Beta Emitter.

Disclaimer, here the m is taken as 0.5kg.

Given, REM = 1500

Time, t= 1 minute =60 seconds

For beta particles,

RBE (Relative Biological Effectiveness) =2.

Absorbed dose,  D =REM/RBE= 1500/2 = 750rad

As 100 rad=1 J/kg,

750 rad=7.5 J/kg

So, D=7.5 J/kg

D= total energy absorbed (E)/mass (m)

7.5 J/kg=E/m

E =7.5 ×0.5 = 3.75J

Therefore, the number of beta particles absorbed is 3.75 J.

The number of beta particles absorbed,

As 1 MeV= 1.6 ×10⁻¹³J

So, converting 1.1 MeV in J, 1.1×1.6 ×10⁻¹³J=1.76×10⁻¹³J

Numbers of beta particles,

n=3.75J/1.76×10⁻¹³J

n=2.13×10⁻¹²

Hence, the number of beta particles is 2.13×10⁻¹².

Activity, A= n/t

where n is the number of beta particles and t is the time.

A= n/t=(2.13×10⁻¹²)/60

A= 3.5 ×10⁻¹⁰ Bq

So, the activity is 3.5 ×10⁻¹⁰ Bq.

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PLEASE HELP!! 50 points. The answer is not 2.6 m/s.

Answers

Answer:

A

Explanation:

5.2mm/s

hope this helps

A rocket is fired with an initial Velocity OF 100m/s at an angle OF 55 above the horizontal, It explodes on the mountain Side 12s after its Firing What is x-and y-coordinates of the rocket relative to its Firing point?​

Answers

Answer:

( 688.29 , 276.66 )   meters

Explanation:

Break into horizontal and vertical components

x = 100 cos 55  = 57.36 m/s      

   the x coordinate is then 12 seconds * 57.36 m/s = 688.29 meters

y coordinate is a bit more complex because of the curved path due to gravity

Original y velocity =  100 sin 55

y = vo t - 1/2 (9.81)(t^2)

y =  100 sin 55  (12)  - 1/2 (9.81) (12)^2

      yf = 276.66   meters

What is the mass of an object if it has 915J of energy when it is 6.50m above the ground?
A.851kg
B0.0696kg
C.583kg
D.14.4kg

Answers

Answer:

D

Explanation:

GPE = mgh

GPE / gh = m

m = GPE/gh

m = 915 / 6.5 x 9.8

m = 915 / 63.7

m = 14.4kg

Which sentence describes an object that has kinetic energy?

Answers

Answer:

Kinetic energy is referred to the energy associated with movement and or motion, thus the only object that is moving would be A boat sailing across the ocean.

Explanation:

How many minutes would be required for a 300.0 W (i.e., power=energy per second=300 J/s)
immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C? (Note: The specific heat of water
is 4184 J/kg°C)

Answers

Hence, the time required is approximately equal to 4 minutes and 39 seconds.

Given, Power, P = 300 W= 300J/s

Power= Energy per second

Mass, m= 0.25 kg

Initial Temperature= 20.0°C

Final Temperature= 100.0°C

Temperature difference, T =( 100-20)°C= 80.0°C

Specific heat of water, S= 4184 J/kg°C

Energy can be represented as

E=mST

where E is the energy, m is the mass, S is the specific heat of water and T is the temperature difference.

As Energy can be written as the product of time and power, E=Pt

Pt=mST

So,

t=(msT)/P

t=(0.25×4184×80)/300

Time, t = 278.934 seconds= 4 minutes and 39 seconds.

Hence, 4 minutes and 39 seconds will be required for a 300.0 W immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C.

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Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform
angular acceleration of 0.01 rad/s² about the centre

Answers

Answer:

r = 2.0m

when t = 0, the angular displacement is zero.

Explanation:

When the angular acceleration is 0.01 m/s2, you can calculate

angular acceleration = change in angular velocity/ time

all elements in the same group have the same number of?
electrons
energy levels
electrons in the outermost energy level
bonds

Answers

All elements in the same group have the same number of electrons in the outermost energy level. Option C is correct.

What is a periodic table?

The periodic table is an organization of all known elements in order of increasing atomic number and repeating chemical characteristics.

They are organized in a tabular format, with a row representing an era and a column representing a group. E elements are ordered in increasing atomic number order from left to right and top to bottom.

Thus, Elements of the same group will have the same valence electron configuration and, as a result, will have comparable chemical characteristics.

Each set of elements has an equal amount of electrons in the outer orbital. Valence electrons are another name for the outer electrons.

They are the electrons interacting with other elements in chemical bonding. Each element in group one, the first column, has one electron in its outer shell.

Hence option C is correct.

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p ⇒ q, r ⇒ s, p ∨r q ∧ s is invalid

Answers

Answer:

My explanation and answer are down here↓:

Explanation:

p→q is false only when p is true and q is false.

∴p→(q∨r) is false when p is true and (q∨r) is false, and

q∨r false when both q,r are false.

Hence T,F,F.

what type of forces that created at shaping machine

Answers

Answer:

a magnetic force

Explanation:

After the terrorist attacks of September 11th, 2001, the National Center on the Psychology of Terrorism was founded to research psychological aspects of terrorism. This is an example of how __________ affect theories and perspectives in psychology.
A.
current events
B.
new discoveries
C.
oppositions
D.
personal ideologies

Answers

reposta letra C oppositions

Answer asap for brainlist

Answers

They use less wire, A

You are standing in an open field near a train track as a train goes by at a constant speed blowing its horn in all directions (except down, into the ground) at a constant frequency. You are first able to hear it when it is approaching you 1 km away. When it is a quarter of a kilometer past you, by how many decibels has the train's sound level intensity shifted?
(options: 6, 12, 58, 105, 108, or 114)

Answers

Answer:

108 is the correct answer

A hollow conducting sphere has an inside radius of r1 = 0.17 m and an outside radius of r2 = 0.28 m. The sphere has a net charge of Q = 4.8E-06 C. What is the magnitude of the field, in newtons per coulomb, at a distance of 0.05 m from the center of the sphere?

Answers

Explanation:

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10 precautions for obtaining the refractive index of a triangular glass prism. ​

Answers

Answer:

Following precautions for obtaining the refractive index of a triangular glass prism:-

Explanation:

(1) All the faces of the prism should be neat and clean.

(2) Pins for holding the paper on the drawing board must be pinned perpendicular to the paper for better handling.

(3) While fixing the pins for checking the refractive index of the prism, make sure that the reflective images of the pins should be aligned to your eye to avoid any type of parallax error.

(4) Pins should not be removed or pinned again during the experiment.

(5) Avoid mishandling of the prism.

(6) Same edge of the prism should be taken as vertex for observations.

(7) For drawing the boundary of the prism, a sharp pencil should be used.

(8) Soft board and pointed pins should be used.

(9) The distance between the pins should be 5 cm or more.

(10) Make sure the glass slab or prism taken are polished and not broken.

Hope this helps!

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