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3 years ago

Difference between MKS system and CGS system

Physics

1 answer:

ivolga24 [154]3 years ago

8 0

Answer:

MKS stands for Meter, Kilogram and second. In this system of unit mass is given in Kilogram, length in meter and time in second. ... CGS system stands for Centimeter- Gram- Second system. In CGS system, length is measured in centimeters mass is measured in grams and time is in seconds.

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3 years ago

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

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2 years ago

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The mass of the fly is $m_f = 11 mg = 11*10^{-3} g = 1.1*10^{-5} \ kg$

The extension of the web is $e= 4.00 \ mm = 0.004 \ m$

The spring constant is mathematically evaluated as

$k = \frac{mg}{e}$

substituting values

$k = \frac{1.1 *10^{-5} *9.8}{0.004}$

$k = 0.027 \ N/m$

The frequency of vibration is

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

substituting values

$f = \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }$

$f = 7.9 Hz$

8 0

3 years ago

What kind of units as force measured by

Readme [11.4K]

Answer:

Newtons

Explanation:

5 0

3 years ago

Read 2 more answers

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

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Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, $\mu=0.39$

To find,

The maximum speed of car.

Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

$v=\sqrt{\mu rg}$

$v=\sqrt{0.39\times 79\times 9.8}$

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

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3 years ago

Difference between MKS system and CGS system​

Difference between MKS system and CGS system