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Romashka-Z-Leto [24]
3 years ago
14

A construction worker runs a jack hammer for 1.5 hours. The jack hammer has 2.4 kilowatts of power. How much electrical energy d

oes the construction worker use running the jack hammer?
A. 3.6 kilowatt-hours

B. 3600 kilowatt-hours

C. 0.625 kilowatt-hours

D. 1.6 kilowatt-hours
Physics
1 answer:
scoray [572]3 years ago
8 0
Its a because 24*15 is 360 but you have to put the decimal back so its 3.6 kilowats per hour plz mark brainlest 
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Answer:

Increase, same, same

Explanation:

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2 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

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\Delta K is the change in kinetic energy

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Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

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6 0
3 years ago
A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h
raketka [301]

Answer:

\rho = 1848.03 kg m^{-3}

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height  of water  = 20 cm  =0.2 m

Pressure  p = 1.01300*10^5 Pa

pressure at bottom

P =  P_{fluid} + P_{h_2 o}

P   = P_{fluid}  + \rho g h

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p_{fluid}  = P_{atm} + \rho g h_{fluid}                       h_[fluid} = 0.307m

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Ig will because it’s the best battery
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Answer:

10 km/h

Explanation:

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