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3 years ago

Differences between looping and simersaulting

1 answer:

3 0

Somersaulting- for longer distances.It bends the narrow end in the direction it wants to go & takes grip with tentacles. It releases the broad end and straightens up. like this it continues. looping- for shorter distances.

Hope this helps

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Answer the question correctly. Look at the 2 pictures.

Alexxandr [17]

Answer:

The answer is 3.15

Explanation:

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2 years ago

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Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the c

son4ous [18]

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?

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3 years ago

Does someone know how to do math with that equation

mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

$\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\$

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

8 0

3 years ago

PLEASE HELP ASAP!!!

musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

$W=Fdcos \theta$

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

$\theta=0^{\circ}$ is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

$W=(20)(1000)=20,000 J$

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

$P=\frac{W}{t}$

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

$P=\frac{20000}{900}=22.2 W$

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0

3 years ago

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

8 0

3 years ago

Differences between looping and simersaulting​

Differences between looping and simersaulting