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andrezito [222]
3 years ago
6

A caterpillar climbs up a one-meter wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the caterpil

lar to climb to the top. Calculate the time involved in seconds (not minutes).
Physics
2 answers:
Brut [27]3 years ago
6 0

100cm to go in 600 secs = 10mins

2 positive, 1 negative .... net 1 positive per step

ipn [44]3 years ago
3 0

Answer:

600 seconds.

Explanation:

A caterpillar climbs up a one meter (100 cm) wall.

For every 2 cm caterpillar climbs up and slides down 1 cm.

That means caterpillar climbs up = 2 - 1 = 1 cm

Caterpillar reaches to the top in 10 minutes. We have to calculate the time in seconds.

Since 1 minute = 60 seconds

Therefore, 10 minutes = 10×60 = 600 seconds

Time involved for the caterpillar to climb to the top will be 600 seconds.

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In which does sound travel fastest? Question 8 options: solids liquids gases plasma.
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<h3>What is sound?</h3>

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Friction is a ____________ force<br> a. Artificial<br> b. Natural<br> c. Pessimistic<br> d. Negative
Daniel [21]

Answer:

natural is the answer

5 0
2 years ago
Read 2 more answers
an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe
yulyashka [42]

Answer:

<em>The amount of electric charge transported = 0.192 C</em>

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

<em>Given:</em> I = 285 mA, t = 674 milliseconds.

<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>

<em>       (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s          </em>

Substituting these values into equation 1

Q = 0.285 × 0.674

<em>Q = 0.192 C</em>

<em>Therefore the amount of electric charge transported = 0.192 C</em>

<em></em>

<em></em>

5 0
2 years ago
car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down ove
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Answer:

78.4 KN/m

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Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'=   1 cm  => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e  F=kx

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k=78.4x10^3

k=78.4 KN/m

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3 years ago
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