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3 years ago

Concept map for kinetic energy, work and power

Physics

1 answer:

Helen [10]3 years ago

8 0

Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:

$W=K_f - K_i$

where

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as

$P=\frac{W}{t}$

where

W is the work done

t is the time elapsed

Learn more about kinetic energy, work and power:

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Name an element in the same period as rubidium

Artyom0805 [142]

Answer: Potassium(K)

Explanation:

its an alkali metal placed under sodium and its over rubidium, its also the first element of period 4

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2 years ago

A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___

kondaur [170]

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3 years ago

Read 2 more answers

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

DerKrebs [107]

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

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3 years ago

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$