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3 years ago

A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?

Physics

2 answers:

Cerrena [4.2K]3 years ago

7 0

Answer:

0,2 would be the coefficient of friction.

Explanation:

Remember that the force necessary to move an object over a horizontal surface is the weight of the object multiplied by the coeficient of friction, an the formula is this:

$Friction= u*N$

Where u is the coefficient of static friction and N is the normal force, or in this case the weight of the object so if we solve for U the formula would be like this:

$u=\frac{Friction}{N}$

Now we just insert the data we know:

$u=\frac{Friction}{N}\\u=\frac{5N}{25N}\\ u=0,2$

So the coefficient of friction would be 0.2

olganol [36]3 years ago

3 0

Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction

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stiks02 [169]

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time taken for the collision

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A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration

Pani-rosa [81]

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

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3 years ago

Read 2 more answers

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

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3 years ago

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3 years ago

A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14

Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

$v_{B} = \sqrt{237.16}$

vB = 15.4 m/s : speed of the cart at B

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3 years ago