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Alex17521 [72]
4 years ago
12

An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s in

terval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Physics
1 answer:
Ronch [10]4 years ago
8 0

Explanation:

A.

Given:

V = 13 m/s

t = 4 s

Constant acceleration, a= (V-Vi)/t

= 13/4

= 3.25 m/s^2

F = mass * acceleration

= 2.7 * 3.25

= 8.775 N.

Using equations of motion,

distance,S = (13 * 4) - (1/2)(3.25)(4^2)

= 26 m

Workdone, W = force * distance

= 8.775 * 26

= 228.15 J

B.

Instantaneous power, P = Force *Velocity

= 8.775 * 13

= 114. 075 W

C.

t = 2 s,

Constant acceleration, a= (V-Vi)/t

= 13/2

= 6.5 m/s^2

Force = mass * acceleration

= 2.7 * 6.5

= 17.55 N

Instantaneous power, P = Force *Velocity

= 17.55 * 13

= 228.15 W.

= 114. 075 W.

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The total units by an objects as it changes position is called _____ ?
Alika [10]
Change in position of object = Displacment
7 0
3 years ago
A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3
Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

\Delta P \propto \rho \cdot \Delta h

\Delta P = k \cdot \rho \cdot \Delta h

Where:

\Delta P - Manometric pressure difference, measured in kilopascals.

\rho - Fluid density, measured in kilograms per cubic meter.

\Delta h - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}

\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}

Where:

\Delta h_{air} - Height difference of the air column, measured in meters.

\Delta h_{Hg} - Height difference of the mercury column, measured in meters.

\rho_{air} - Density of air, measured in kilograms per cubic meter.

\rho_{Hg} - Density of mercury, measured in kilograms per cubic meter.

If \Delta h_{Hg} = 0.015\,m, \rho_{air} = 1.29\,\frac{kg}{m^{3}} and \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, the height difference of the air column is:

\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)

\Delta h_{air} = 158.140\,m

The height of the building is 158.140 meters.

5 0
3 years ago
Read 2 more answers
The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 k
LuckyWell [14K]

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3)(90)(4.44)^2 = 0.33 x 90 x 19.7136 = 585.49

w= 1 rev/hour = 1 rev/3600sec = 2pi/ 3600 = pi/1800 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/1800)^2 = 0.5 x 0.33 x 90 x 19.7136 x 3.05 x 10^-6

KE = 0.00089 J

For hour hand

I = (1/3)(58.2)(2.79)^2 = 0.33 x 58.2 x 2.79^2 = 149.5

w = 1 rev/12hour = 1 rev/(12x3600sec) = 2pi/ 12x3600 = pi/21600 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/21600)^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J

4 0
4 years ago
A 500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the
malfutka [58]

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

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