9. To completely describe the motion of an object, you need

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

- When an object is dropped what happens to potential and kinetic energy as it falls?

aev [14]

The answer is D: the GPE goes down and the KE goes up

7 0

3 years ago

A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a

AysviL [449]

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

$EMF = \frac{\phi_2 - \phi_1}{\Delta t}$

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

$\phi_1 = NBAcos 0 = BA$

$\phi_2 = NBAcos 90 = 0$

now we will have

$EMF = \frac{NBA}{t}$

$10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}$

$B = 0.126 T$

3 0

3 years ago

Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively, kept at a distance of 10 m from

NNADVOKAT [17]

Answer:

Answer

F=G×

d

2

m×M

m = 50 kg

M = 120 kg

Distance, d = 10 m

G=6.7×10

−11

Nm

2

/kg

2

F=6.67×10

−11

×

10

2

50×120

F=6.67×60×10

−11

F=4.02×10

9

N

6 0

3 years ago

Read 2 more answers

Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed

quester [9]

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

q₀ = 4.0 10⁻⁶ C

Let's calculate

F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

F = - 1,598 10⁻³ N

4 0

4 years ago