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Elis [28]
3 years ago
6

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Physics
1 answer:
Elenna [48]3 years ago
3 0

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

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Answer:

73.72

Explanation:

For this subtraction problem, the answer or solution is expressed to the least precise of the numbers we are trying to subtract.

The least precise number is the number with the lowest significant numbers:

105.4 - 31.681

105.4  has 4 significant numbers

31.681 has 5 significant numbers

  So;

             105.4  

        -      31.681

        ------------------

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The solution is therefore  73.72

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3 years ago
build three straight ramps on the stage one ramp should be nearly flat and another should be extremely steep the last ramp shoul
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Answer:

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Explanation:

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1 year ago
A 4.0 kg mass is attached to a spring whose spring constant is 950 N/m. It oscillates with an amplitude of 0.12 m. What is the m
Fudgin [204]

Answer:

velocity = 2.62m/s

Explanation:

950= (4 x A)/0.12

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A = 28.5m/s²

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When you eat a large meal and your body absorbs a lot of glucose and that makes its way to the interstitial fluid before going i
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The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general
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Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

O=(0,0)

A=(0,-b)

B=(h,0)

We need to calculate the position vector of AB

\bar{AB}=(h-0)i+(0-(-b))j

\bar{AB}=hi+bj

We need to calculate the unit vector along AB

u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}

u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}

We need to calculate the force acting along the edge

\hat{F}=F(u_{AB})

\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})

We need to calculate the net moment

\hat{M}=\hat{F}\times OA

Put the value into the formula

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\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))

\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})

\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}

Put the value into the formula

\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}

\hat{M}=-81.102\ \hat{k}\ N-m

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0
3 years ago
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