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3 years ago

What tool would you use to separate nails from gravel

Physics

2 answers:

Minchanka [31]3 years ago

7 0

It would be a wire mesh

nekit [7.7K]3 years ago

3 0

I think a magnet would be best. It's cheap, quick and easy,
and very technical, so that's the one I would use.

You might be interested in

Look around the room you are in. Name

kozerog [31]

Answer:

A wire,rubber

Explanation:

yes

3 0

3 years ago

Read 2 more answers

You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. I

DochEvi [55]

Answer:

Explanation:

The net force on the potatoes is given by:

F= 52 - mgSintheta

F= 52- (2×9.8× Sin70°)

F = 52 -18.4

F= 33.58N

Using Newton's 2nd law

F = ma

a=F/m = 33.58/ 2 = 16.79m/s^2

Using the equation of motion:

V^2= u^2 + 2as

V^2 = 0 + 2× 16.79 x2

V^2 = 67.16

V=sqrt(68.16)

V= 8.195m/s This is the exit velocity of the potatoes

Kinetic energy, K.E = 1/2mv^2

KE= 1/2 × 2 × 8.195^2

KE = 67.16J

8 0

3 years ago

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

1,230N

Explanation:

1. Name of the variables:

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

2. Formulae:

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

3. Solution (calculations)

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

Reading or writing on a phone is especially dangerous while driving because:

algol [13]

Answer:

Explanation:

When we are driving we need a lot of attention and concentration. Also one involved in driving should be consious and courteous

Thus, whenever a person is drives, and when he is disactracted by Mobile phones it will destroy his presence of mind.

It will good if use mobile after stopping the vehicle

Thanks

5 0

3 years ago

Read 2 more answers