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Anon25 [30]
2 years ago
7

Brutus, a champion weight lifter, raises 220 kg a distance of 3.10 m. (a) how much work is done by brutus lifting the weights? j

(b) how much work is done holding the weights above his head? j (c) how much work is done lowering them back to the ground? j(d) does brutus do work if the weights are let go and fall back to the ground? no yes (e) if brutus completes the lift in 2.1 s, how much power is developed? w
Physics
1 answer:
Nana76 [90]2 years ago
3 0
A) work = force * distance 
mass is not a force, weight is, so we have to find the weight of the block. 
Weight = mg 
Weight = (220kg)(9.8) 
Weight = 2156N 
Work = 2156N * 3.10m 
work = 6683.6J 
b) Since he is holding the weights, it's not moving, therefore, he doesn't do any work 
c) The answer is still the same amount of work when he lifted them. 
d) The answer is no since when he let go the weight, he doesn't apply any force to the weight. 
e) P = work/time 
P = 6683.6J / 2.1s 
P = 3182.67 watts 
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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
How is force related to math
dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

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2 years ago
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How do heat insulators work
a_sh-v [17]
The heat is transferred to one material to another, however insulators minimize that transfer, keeping it in the area, warming it. 
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What would happen if you drop and egg
dexar [7]

Answer:

It would crack.

Explanation: The pressure from dropping it would crush the eggshell therefore breaking the egg.

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