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-Dominant- [34]
2 years ago
11

How do objects and substances in motion have kinetic energy

Physics
1 answer:
Fiesta28 [93]2 years ago
6 0

Answer:

Explanation:

Kinetic energy is one of the types of mechanical energy and it is the energy due to the object in motion.

Kinetic energy is given as 1/2mv^2

A body cannot have a K.E when it is not in motion because of the velocity

Velocity is the change is position of an object or particle with time taken, so if the object doesn't change it position then it won't have velocity.

kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

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A sailboat weighing 980 lb with its occupants is running downwind at 8 mi/h when its spinnaker is raised to increase its speed.
Effectus [21]

Answer:

78.498N

Explanation:

The Net force provided by the spinnaker can be obtained from Newton's second law of motion as follows;

F=\frac{m(v-u)}{t}................(1)

where m is the mass, v is the final velocity, u is the initial velocity and t is the time interval for which the force acted.

Given;

m =980lb

v = 12mi/h

u =8mi/hr

t = 10s.

It is important to convert all quantities to their SI units where necessary, so we do that as follows;

1lb = 0.45kg,

hence 980lb = 980 x 0.45kg = 441kg.

1mile = 1609.34m

1hour = 3600s,

therefore;

8mi/h=\frac{8*1609.34m}{3600s}=3.58m/s

12mi/h=\frac{12*1609.34m}{3600s}=5.36m/s

Substituting all values into equation (1), we obtain the following;

F=\frac{441(5.36-3.58)}{10}\\F=\frac{441*1.78}{10}\\F=\frac{784.98}{10}\\F=78.498N

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Most workers in nanotechnology are actively monitored for excess static charge buildup. the human body acts like an insulator as
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2 years ago
A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
3 years ago
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