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tresset_1 [31]
2 years ago
9

Agatha the snake is 50 centimeters long.3 years from now she will be 152 centimeters long and fully grown. If Agatha grows at co

nstant rate, how much does she grow in length each year​
Physics
1 answer:
PIT_PIT [208]2 years ago
8 0

Answer:  34cm each year

Explanation:

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What are basic physical quantities and it's measurement units? Make a chart.
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8 0
2 years ago
When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin
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True
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7 0
2 years ago
Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a
FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

1* t - 2*(t - 120) = 10

t - 2t + 240 = 10

t = 230 s

now the distance moved by Blaise is given by

d_1 = 1*230 = 230 m

6 0
2 years ago
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 50n box is lifted 2 meters in 3 seconds.
anygoal [31]

F=G=50 N

L=F*d=50*2=100 J

is the work

P=L/t=100/3=33.33 W

4 0
3 years ago
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