A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1

Answers

Answer 1

Answer:

The correct option is  B

Explanation:

From the question we are told that

   The mass of the pile is  M

   The  height is  H  =  4 y

    The vertical distance achieve during the first lift is  [tex]h_1  =  3 y[/tex]

     The time taken is  [tex]t_1 =  4T [/tex]

    The vertical distance achieve during the second lift is  [tex]h_2  =  y[/tex]

     The time taken is [tex] t_2 =  T [/tex]

Generally the velocity of the crane during the first lift is  

      [tex]v _1 =  \frac{h_1}{t_1 }[/tex]

=>    [tex]v _1 =  \frac{3 y}{4T }[/tex]    

Generally the velocity of the crane during the second  lift is    

      [tex]v _2 =  \frac{h_2}{t_2 }[/tex]

=>    [tex]v _2 =  \frac{ y}{T}[/tex]  

Generally the power generated by  the crane during the first lift is    

    [tex]P_1 =  F_1 *  v_1[/tex]

Here [tex]F_1[/tex] is the weight of the brick which is mathematically represented as

      [tex]F_1 =  M  * g [/tex] , g is the acceleration due to gravity

 So

       [tex]P_1 =  Mg  *  \frac{3y}{4T}[/tex]

Generally the power generated by  the crane during the first lift is    

    [tex]P_1 =  F_2 *  v_2[/tex]

Here [tex]F_2[/tex] is the weight of the brick which is mathematically represented as

      [tex]F_2 =  M  * g [/tex] , g is the acceleration due to gravity

 So

       [tex]P_1 =  Mg  *  \frac{y}{T}[/tex]

The ratio of the first power generated to the second power is  

       [tex]\frac{P_1}{P_2} =  \frac{Mg  *  \frac{3y}{4T} }{ Mg  *  \frac{y}{T} }[/tex]

=>    [tex]\frac{P_1}{P_2}  =  \frac{3}{4}[/tex]

=>   [tex]P_2 = \frac{4}{3} P_1[/tex]


Related Questions

Plates slide past one another at____.
A. Subduction zones
B. Transform boundaries
C. Convection currents
D. Divergent boundaries

Answers

Answer:

Transform Boundary

Explanation:

The just slide past each other

Answer:

Transform Boundaries

Explanation:

1. What is Ohm"s law?
2. If you placed a negatively charged hairbrush near your hair, what charge would your hair be?
3. You must change a lightbulb and the new lightbulb has a larger resistance. If the voltage of the battery does not change, what happens to the current going through the flashlight?
HELLPPPP

Answers

1. Ohm's law shows the relationship between:

voltagecurrentresistance

Formula: voltage = current x resistance

2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)

3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.

Best of Regards!

A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.

Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.

Answers

Answer:

The value  is    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

Explanation:

From the question we are told that

  The atom fraction of metal A at point G is [tex] A  =  0.30 \ m[/tex]

   The atom fraction of metal  A at a distance 5000nm from G is  [tex]A_2 = 0.35[/tex]

   The number of atoms per [tex]m^3[/tex] is    [tex]N_h =  9 * 10^{28}[/tex]

    The diffusion coefficient is  [tex]D =   2* 10^{-14 } m^2/s[/tex]

Generally of the concentration of atoms of metal A at G is  

       [tex] N_A = A * N_h [/tex]

=>    [tex] N_A =  0.3  * 9 * 10^{28}[/tex]

=>     [tex] N_A =   2.7 * 10^{28} 2.7 atoms/m^3[/tex]

Generally of the concentration of atoms of metal A at a distance 5000nm from G is  

       [tex]D =  0.35 *9 * 10^{28}[/tex]

=>     [tex]D =  3.15 * 10^{28} \  atoms / m^3[/tex]

The concentration gradient is mathematically represented as

   [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]

=> [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]  

=>   [tex]\frac{dN_A}{dx}  = 9 * 10^{20} / m^4[/tex]  

Generally the flux of the atoms per unit  area according to Fick's Law  is mathematically represented as

       [tex]J =  -D* \frac{d N_A}{dx}[/tex]

=>    [tex]J =  -2* 10^{-14 * 9 * 10^{20} [/tex]

=>    [tex] J =  18*10^{6}\   atoms\ crossing\ /m^2 s  [/tex]

Generally if the cross-section area is [tex] a  =  1 cm^2 =  10^{-4} \  m^2[/tex]

Generally the number of atom crossing the above area  per second is mathematically is  

      [tex]H  =  18*10^{6}    *  10^{-4} [/tex]

=>    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

1. What is the chemical name of the following chemical symbols?
Be
Nel
Mg
Na

Answers

Answer:

Na sodium

Mg magnesium

Be beryllium

Explanation:

Nel is not any element it is wrong

A metal ball sits motionless on a flat surface. Which of these would make the ball move?
A. The force of gravity becomes less.
B. The force of gravity becomes greater.
C. Two equal horizontal opposing forces act upon the ball.
D. Two unequal horizontal opposing forces act upon the ball.

Answers

Answer:

D

Explanation:

Unbalanced forces move stuff. Gravity would only increase/decrease movement if the object was already in motion.

Answer:

b

Explanation:

The volume of water in a water bottle, is about 398
g
cm
km/hr
Kg
g/mL
ml
km
m/s

Answers

Answer:

milliliters (ml)

Explanation:

millileters is the correct measurement for liquids

calculating light in physics

Answers

Formula: c = f where:
c = the speed of light = 300,000 km/s or 3.0 x 108 m/s.
= the wavelength of light, usually measured in meters or Ångströms (1 Å = 10-10 m)
f = the frequency at which light waves pass by, measured in units of per seconds (1/s).

1
2
3
4
5
6
8
9
10
Dressing appropriately for exercise includes
A. wearing the same clothes for all exercises
B. choosing dark colored clothing when exercising at night
C. wearing sunscreen when exercising outside
D. making sure you wear the best brand-name clothes
Please select the best answer from the choices provided.
A
B.
C.
D.

The answer is C.

Answers

Answer:

the answer is C

Explanation:

you said its c

In which medium does the light move faster, water or diamond?

Answers

Answer:Light moves faster in after than that of diamonds

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 7.63 m/s in 3.94 s. What is the magnitude of the linear impulse experienced by a 73.7 kg passenger in the car during this time? Submit Answer Tries 0/20 What is the average force experienced by the passenger?

Answers

Answer:

1. p = 562.3 kg*m/s

2. F = 142.7 N

Explanation:

1. The linear impulse (p) is given by:

[tex] p = mv [/tex]

Where:

m: is the passenger's mass = 73.7 kg

v: is the speed = 7.63 m/s

[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]

Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.

2. The average force can be calculated using the following equation:

[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]  

Therefore, the average force experienced by the passenger is 142.7 N.

I hope it helps you!

An object is accelerating if it is moving?


Answers

9514 1404 393

Answer:

  Not Necessarily

Explanation:

If the object is changing speed or direction, then it is accelerating. If it is maintaining the same speed and direction, it is not accelerating.

At time t = 0 the point at x = 0 has velocity v0 and displacement y0. The phase constant φ is given by tanφ =:

Answers

This question is incomplete, the complete question is;

The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .

At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.

The phase constant φ is given by tanφ =:

A) ωv₀ /y₀    

B) ωv₀ y₀  

C) v₀ /ωy₀  

D) y₀ /ωv₀      

E) ωy₀ /v₀

Answer:

E) ωy₀ /v₀

Explanation:

Given that;

displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)

we differentiate the given equation with respect to time

d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )

v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )

v₀ = -ym ωcos (-φ)  ......... lets leave thisas equ 1

At t = 0, x = 0

the displacement of the wave is

y(0,0) = ym sin (k(0) - ω(0) - φ)

y₀ = ym sin(-φ) ..............let this be equ 2

y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))

(tanφ)/ω = y₀/v₀

tanφ = y₀ω/v₀

therefore the required value is y₀ω/v₀

option (E).  

The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be

Answers

This question is incomplete

Complete Question

m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.

a) if the system is released from rest what will be its acceleration

Answer:

0.7 m/s²

Explanation:

The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.

(a) if the system is released from rest what will be its acceleration

g = acceleration due to gravity = 9.81 m/s²

Coefficient of Kinetic Friction = μk = 0.30

m1 = 10kg

m2 = 4.0kg

The formula to solve question a is given as:

a = acceleration at rest

m2g- μk m1g = (m1+ m2) a

Making a the subject of the formula:

a = (m2g- μk×m1g )/(m1+ m2)

a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)

a = 0.7 m/s²

if you are driving 110 km/h along a straight road and you look to the side for 2.0 s , how far do you travel during this inattentive period ? explain. ​

Answers

Explanation:

hope this helps, have a good one :D

Answer:

60.12m

Explanation:

Distance = Velocity x Time

To use this formula we must first convert 110km/h to m/s, which we can do by dividing the value by 3.6:

110/3.6 = 30.56m/s (2dp)

Velocity = 30.56m/s

Time = 2s

Distance = 30.56x2

Distance = 61.12m

You travel 60.12m during this inattentive period.

Hope this helped!

what happens to the temperature of water as time elapses? IF YOU ANSWER IT I WILL MARK YOU A BRAINLEST ANSWER​

Answers

Answer:

I think it will get colder

Explanation:

Answer:

The water molecules go faster as it gets colder they go slower

Explanation:

trust me thats the answer

If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m​

Answers

Answer:

(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Explanation:

Vectors

Given a vector

[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]

[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]

[tex]\mid A\mid=7[/tex]

The unit vector is:

[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]

[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]

[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]

[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Find the angle between the two unitless vectors: F1 = 8.92 i + 17.37 j F2 = 12.44 i + 7.11 j Answer in degrees, and to the fourth decimal place.

Answers

Answer:

θ = 33.0705°

Explanation:

The angle between the two vectors is given by the formula;

Cos θ = (F1 • F2)/(|F1| × |F2|)

We are given;

F1 = 8.92i + 17.37j

F2 = 12.44i + 7.11j

Thus;

Cos θ = [(8.92i + 17.37j) • (12.44i + 7.11j)]/[√(8.92² + 17.37²) × √(12.44² + 7.11²)]

Cos θ = (110.9648 + 123.5007)/(19.5265 × 14.3285)

Cos θ = 0.8380

θ = cos^(-1) 0.8380

θ = 33.0705°

Please provide explanation!!!
Thank you.

Answers

Answer:

(a) 102 cm/s

(b) 0.490 cm²

Explanation:

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what
is the value of the current?
A. 3.0 A
B 5.0 A
C. 11 A
D. 15 A
E. 20A

Answers

Answer:

B 5.0 A .

Explanation:

Hello.

In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:

[tex]I=\frac{Q}{t}[/tex]

Then, for the given data, we obtain:

[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]

Therefore, answer is B 5.0 A .

Best regards!

3 For this force system the equivalent system at P is ___________ A FRP 40 lb along x dir and MRP 60 ft lbB FRP 0 lb and MRP 30 ft lbC FRP 30 lb along y dir and MRP 30 ft lbS FRP 40 lb along x dir and MRP 30 ft lb

Answers

This question is incomplete, the complete question is;

For this force system the equivalent system at P is ___________

A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft.lb

B) FRP = 0 lb and MRP = +30 ft.lb

C) FRP 30 lb (along +y-dir.) and MRP  = -30 ft.lb

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

Answer:

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

Explanation:

From the figure in the image i uploaded along this answer;

FRP = ( 40 lb i + 30 lb j ) + [30 lb (-j)]

Where i  and j are the unit vectors along X & Y axis respectively.

So, FRP = 40 lb i

that is,  FRP = 40 lb along +X direction

MRP = [ 30 lb x ( 1 ' + 1' ) ] +( -30 lb x 1 ' )

= (30 lb x 2 ' )- 30 lb ft

= 60 lb ft - 30 lb ft

= 30 lb ft

Therefore option(D) is correct    

21. A toy car starts from rest and begins to accelerate at 11.0 m/s2. What is the toy
car's final velocity after 6.0 seconds?

Answers

Answer:

Explanation:

Given parameters:

Initial velocity = 0

Acceleration = 11m/s²

Time  = 6s

Unknown:

Final velocity  = ?

Solution:

 From the given parameters, we use one of the appropriate equations of motion to solve this problem.

     V = U + at

V is the final velocity

U is the initial velocity

a is the acceleration due to gravity

t is the time taken

Input the parameters and solve;

     V  = 0 + 11 x6

     V  = 66m/s

The final velocity is 66m/s

I WILL GIVE BRAINLIEST
In which of the following locations would most likely find parenchyma cells? Leaves roots flowers bark

Answers

Answer:

I would guess its leaves

Answer:

Leaves

Explanation:

which equation should be used to find speed

Answers

Answer:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).

An object is rolled at 12 m/s down a table. It stops
after 15s. What was its acceleration?
Variables:
Equation and Solve:

Answers

Answer:

We are given:

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

Solving for acceleration:

from the first equation of motion

v = u + at

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

a = -4 / 5 m/s²

A charged isolated metal sphere of diameter 12 cm has a potential of 9200 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.

Answers

Answer:

0.1 J/m³

Explanation:

We know that

V = k Q / R

We also know that

E = k Q / R²

Joining the two equations together, we have

E = V / R

To solve the question proper, we'd be using the formula

u = 1/2 E• E², substitute for E, we have

u = 1/2 E• (V/R)²

u = 1/2 * 8.85*10^-12 * (9000 / 0.06)²

u = 1/2 * 8.85*10^-12 * 150000²

u = 1/2 * 8.85*10^-12 * 2.25*10^10

u = 1/2 * 0.199125

u = 0.0996

u = 0.1 J/m³

The energy density is 0.1 J/m³

A child and sled with a combined mass of 53.9 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 5.71 m/s at the bottom, what is the height of the hill

Answers

Answer:

1.66m

Explanation:

Using the conservation law

PE = KE

mgh = 1/2mv²

gh = V²/2

g is the acceleration due to gravity = 9.81m/s²

h is the height of the hill

V is the velocity = 5.71m/s

Substitute

9.81h = 5.71²/2

Cross multiply

2×9.81h = 5.71²

19.62h = 32.6041

h = 32.6041/19.62

h = 1.66m

Hence the height of the hill is 1.66m

The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb = 3vw. The student is going to travel a distance d. The energy the student uses when walking is Ew. The energy the student uses when biking is Eb. The ratio EwEb is

Answers

Answer:

3

Explanation:

Can someone please explain how to find the acceleration of the hanging mass?

Answers

Answer:

Acceleration = m/s²

Explanation:

T= Newtons compared to the weight W = Newtons for the hanging mass. If the weight of the hanging mass is less than the frictional resistance force acting on the mass on the table, then the acceleration will be zero.

gold has a density of 19.32g/cm3. if you have a 25 cm3 sample of gold what is the mass of the sample​

Answers

Answer:

ggggggggggggggggggggggggggggg

Explanation:

Answer:

The volume of the sample of gold is

16.51 [tex]cm^{3}[/tex]

Explanation:

The formula for density is:

D= [tex]\frac{M}{V}[/tex].

where:

D is density,

M is mass, and

V is volume.

Rearrange the density formula to isolate volume.

V= [tex]\frac{M}{D}[/tex]

V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]

V= 318.97∅ ×  [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.

V= 16.51 cm³ Au.

anyone to assist me on it ...especial page7 and 8

Answers

Answer:

i needed points it was an emergency sorry

Explanation:

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