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Maslowich
3 years ago
9

a 7 kilogram cat is resting on top of a bookshelf that is 2 meters high. what is the cat’s gravitational potential energy relati

ve to the floor? joules
Physics
2 answers:
viva [34]3 years ago
7 0

Answer:

The cat’s gravitational potential energy relative to the floor is 137.2 Joules.

Explanation:

Given that,

Mass of cat = 7 kg

Height of bookshelf = 2 m

The gravitational potential energy :

The gravitational potential energy is the product of the mass of the object, gravity and height of the object from the ground.

The gravitational potential energy relative to the floor is

P.E = mgh-0

P.E = 7\times9.8\times2-0

P.E=137.2\ J

Hence, The cat’s gravitational potential energy relative to the floor is 137.2 Joules.

attashe74 [19]3 years ago
5 0
PE = mgh
      Mass, m = 7kg, g ≈ 10 m/s², height = 2m     
 
      = 7*10*2
 
       = 140 Joules.


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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

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Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

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\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

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<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

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Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

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saul85 [17]

Answer:

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From the question we are told that  

    The force is  F = 187 \ lb

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     The angle between the resultant force and the first force \theta _1  = 29 ^o 1 ' = 29 + \frac{1}{60}  = 29.0167^o

For us to solve problem we are going to assume that

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According to Sine rule

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According to cosine rule

       R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }

Substituting values

     R = \sqrt{187^2 + 129.9 ^2  + 2 (187 ) (129.9) cos (73.6)}

     R = 256.047 N

 

3 0
3 years ago
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