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Dmitry [639]
3 years ago
14

) you carry a 7.0 kg bag of groceries 1.2 m above the ground at constant velocity across a 2.7 m room. how much work do you do o

n the bag in the process
Physics
1 answer:
lapo4ka [179]3 years ago
3 0
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries

The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.

At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
    = (7.0 kg)*(9.8 m/s²)*(1.2 m)
    = 82.32 J

The change in PE is equal to the work done.

Answer: 82.3 J

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True or false it is impossible to determine weather you are moving unless you can touch another object
Lerok [7]

Answer: false

Explanation:

4 0
3 years ago
Read 2 more answers
As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres
Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

Explanation:

From first Law of Thermodynamics, we know that:

ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

<u>Vi = 0.055 m³ = 55 L</u>

7 0
4 years ago
A pendulum has a length of 2 m and a 30 kg mass hanging on the end. What is the period of the
anastassius [24]

Answer:

T = 2.83701481512 seconds

Explanation:

Hi!

The formula that you will want to use to solve this question is:

T = 2\pi *\sqrt{\frac{L}{g}  }

T--> period

L --> length of the pendulum

g --> acceleration due to gravity (9.8m/s^2)

since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)

so now we can plug in our given info into the formula above and solve!

T = 2*pi * sqrt(2/9.8)

T = 2.83701481512 seconds

*Note*

- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different

I hope this helped!

7 0
3 years ago
Weekend A<br> Assignment<br> Differentiate between forced and damped oscillation
4vir4ik [10]

Answer:

A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.

Explanation:

Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.

<h2><em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u> </u></em></h2>

<em><u>Welcome</u></em><em><u> </u></em>

6 0
3 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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