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2 years ago

In handball, once a player possesses the ball,

Physics

1 answer:

Evgesh-ka [11]2 years ago

3 0

Answer:

If possessing the ball, players must dribble (not similar to a basketball dribble), or can take up to three steps for up to three seconds at a time without dribbling. No attacking or defending players other than the defending goalkeeper are allowed to touch the floor of the goal area (within six metres of the goal).

Explanation:

<em>I HOPE THIS HELPS*</em>

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FREE BRAINLIEST! if you can answer this correctly ill give you brainliest and answer some of the questions you have posted :) th

Gala2k [10]

b) accelerate to the left as much more pressure is pulling it in that direction and on the right however , there is less force .

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2 years ago

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

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2 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

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2 years ago

Which is one way scientists indicate how precise and accurate there experimental measurements are

kherson [118]

They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.

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3 years ago

I need help with #4 and #5?

algol [13]

Explanation:

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2 years ago