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butalik [34]
2 years ago
8

Pls help me i have to submit this due tomorrow :'D

Physics
1 answer:
aniked [119]2 years ago
3 0

\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}

b) Finding total distance :

Distance travelled from 0 s to 5 s :

  • 1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

  • 5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

  • 150 + 1/2 x 5 x 5
  • 150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

  • 36 x 5 + 1/2 x 5 x 4
  • 180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

  • 45 x 5 + 1/2 x 5 x 5
  • 225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

  • 40 x 5 + 1/2 x 10 x 5
  • 200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

  • 20 x 5 + 1/2 x 20 x 5
  • 100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

  • 1/2 x 20 x 5
  • 50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

  • 36/15
  • 12/5
  • 2.4 m/s

d) average velocity

  • 0 m/s (as displacement is equal to 0)

e) average speed

  • 1240/40
  • 31 m/s

f) Part d uses displacement whereas part e uses distance

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Nonamiya [84]

Answer:

1.52 m

Explanation:

We are given that

Maximum force=F=679 N

Mass of rock ,m=385 kg

Distance,d=0.233 m

We have to find the  minimum total length L of the rod required to move the rock.

Torque on rock=T_1=mgd=385\times 9.8\times 0.233=879.11 N

Where g=9.8m/s^2

Torque on man=T_2=Force\times distance=(L-d)\times 679=(L-0.233)\times 679

T_1=T_2

(L-0.233)\times 679=879.11

L-0.233=\frac{879.11}{679}=1.29

L=1.29+0.233=1.523\approx 1.52 m

Hence, the minimum total  length of rod=1.52 m

3 0
3 years ago
15. A girl pushes a box that has a mass of 85 kg up an incline. If the giri exerts a force
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<h3><u>Answer;</u></h3>

M.A = 4.25

<h3><u>Explanation;</u></h3>

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It is the ratio of output force to the input force.

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A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
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Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
3 years ago
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Answer:

wood

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you can last longer with wood because you are using up gas when you drive your car that will mean that you will have to go and find gas if you run out an example like we are in the zombie apocalypse.

8 0
2 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
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Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

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Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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