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lutik1710 [3]
1 year ago
8

Gibbons move through the trees by swinging from successive handholds, as we have seen. To increase their speed, gibbons may brin

g their legs close to their bodies. How does this help them move more quickly?
Physics
1 answer:
Sedbober [7]1 year ago
5 0

By bring their legs close to their bodies, they are decreasing the length of the pendulum which help them move more quickly.

A pendulum is nothing but a body suspended from a fixed point so that it can swing back and forth under the influence of gravity.

Here in this case Gibbons are bringing their legs close to their bodies and reducing the length of the pendulum. Since as the length of the pendulum increases the speed of the movement will be reduced. By bringing their legs close to their bodies they are reducing the length and in turn their speed increase and they move quickly.

To learn more about pendulum, visit here

brainly.com/question/14759840

#SPJ4

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Answer:

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Explanation:

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4 0
3 years ago
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NNADVOKAT [17]

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The correct answer is  "Some substances must be dissolved in water before they can be used".  

5 0
3 years ago
A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0
3 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t
kkurt [141]
Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s

t=12s
4 0
3 years ago
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