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Assoli18 [71]
3 years ago
5

Lee pushes horizontally with a force of 75 n on a 36 kg mass for 10 m across a floor. calculate the amount of work lee did. answ

er in units of j.
Physics
2 answers:
svlad2 [7]3 years ago
8 0
To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.

So you would do 75 N x 10m x Cos (0 degrees)= 750 J
Marysya12 [62]3 years ago
7 0

Answer:

The work done by Lee is 3528 Joules.

Explanation:

Given that,

Mass of the object, m = 36 kg

Force applied by Lee on the object, F = 75 N

The object moves to a distance, d = 10 m

We need to find the amount of work done by Lee. It is given by the combination of force and displacement of an object. It is given by :

W=F{\cdot} d

W=mg{\cdot} d

W=36\times 9.8\times 10

W = 3528 Joules

So, the work done by Lee is 3528 Joules. Hence, this is the required solution.

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Bulky is working at a UPS loading dock. He boasts he can lift a 154 kg box 4 meters in 30
ivanzaharov [21]

Answer: Power is 200 W

Explanation: Power P = work done / time used.

P = W/t = mgh/t = 154 kg · 9.81 m/s²· 4 m / 30 s = 201 W

8 0
3 years ago
Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system?
givi [52]
Thermal energy from the coffee is transferred to the mug.
3 0
3 years ago
A spaceship whose rest length is 350m has a speed of .82c
igomit [66]

Answer:

t'=1.1897*10^{-6} s

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}

where:

v is the velocity of spaceship relative to certain frame of reference =  -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }

u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

t'=\frac{length}{Relatie seed (u')}

t'=\frac{350}{0.9806c}     (c = 3*10^8 m/s)

t'=\frac{350}{0.9806* 3.0*10^{8} }

t'=1.1897*10^{-6} s

t'=1.1897 μs

4 0
3 years ago
a piece of metal with a mass of 15.3 grams has a temperature of 50.0°C. When the metal is placed in 80.2 grams of water at 21.0°
AleksAgata [21]

Answer:

1.21

Explanation:

Heat rise in the body happens due to heat supplied by water to the body.

Heat rise in body = m₁ c₁ ΔT₁

Where m₁ is mass of body and c₁ is its specific heat of body

Heat lost from water to the body = m₂ c₂ ΔT₂

Where m₂ is mass of water and c₂ is its specific heat of water ( c₂ =1 (since water))

Equating both:

        15.3 x c₁ x 4.3 = 80.2 x 1 x 4.3

⇒   c₁ = 80.2 / (15.3 x 4.3) = 1.21

6 0
3 years ago
Read 2 more answers
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
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