Giving 25 points need help ASAP!

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0

3 years ago

Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l

wlad13 [49]

The solution would be like this for this specific problem:

The force on m is:

GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:

GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2

Since the Ms are the same, then

2mx^2 / L^2 = m(L - x)^2 / L^2

2x^2 = (L - x)^2

xsqrt2 = L - x

x(1 + sqrt2) = L

x = L / (sqrt2 + 1) From here, we rationalize.

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)

x = L(sqrt2 - 1) / (2 - 1)

x = L(sqrt2 - 1)

= 0.414L

Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.

8 0

3 years ago

A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric

MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

$J = \frac{I}{\pi R^2}$

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

$J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}$

This enclosed current is now to be used in Ampere’s Law.

$\mu_o I_{enc} = \int {B} \, dl$

Here, $\int \, dl$ represents the circular path of radius r. So we can replace the integral with the circumference of the path, $2\pi r$ .

As a result, the magnetic field is

$B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}$

5 0

3 years ago

Why is it very difficult for astronomers to find objects like planets and asteroids?

mezya [45]

It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location

3 0

3 years ago

You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas

sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 $\frac{a}{g}=\frac{25}{196}$

$a=1.2512m/sec^2$

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m$

7 0

3 years ago