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gregori [183]
3 years ago
10

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates

. A 72.0-kg water-skier has an initial speed of 5.10 m/s. Later, while moving along a straight line, the speed increases to 12.1 m/s. Determine the work done by the net external force acting on the skier.
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

4334.4 J

Explanation:

Work done equals to kinetic energy change

KE=½mv²

Change in KE is given by

∆KE=½m(v²-u²)

Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.

Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then

∆KE=½*72(12.1²-5.10²)=4334.4J

Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J

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The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space
Svet_ta [14]
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4 0
3 years ago
If i apply 280 n of force to a 40kg object, what will it's acceleration be?
nika2105 [10]

Answer:

f=ma

f=280N

m=40kg

a=?

280=40a

a=280/40

a=70N/kg

6 0
3 years ago
How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences​
aliina [53]

Answer:

The lungs get rid of carbon dioxide and water vapor. The liver gets rid of bile, which, in addition to breaking down fats, is partially made up of the breakdown of red blood cells. The kidneys get rid of toxins from the blood. The large intestine gets rid of undigested food

3 0
3 years ago
One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau
suter [353]

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0
3 years ago
1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown
denis23 [38]

Answer:

a) 3.37 x 10^{3} kg/m^3

b) 6.42kg/m^{3}

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x 10^{-4}  kg/m^{3}. So density of metal = mass of metal / volume of metal = 5 / 14 x 10^{-4}  kg/m^{3} = 3.37 x 10^{3} kg/m^3

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/m^{3}

3 0
3 years ago
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