PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! The two main types of circuits are:

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago

Que es la expansión del universo?

anastassius [24]

Answer:

La expansión no es más que el incremento con el tiempo de la distancia entre cualquier par de galaxias lejanas. Se suele utilizar para representar este hecho la analogía de un globo donde hemos pintado una serie de puntos a modo de galaxias.

Explanation:

3 0

2 years ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

2 years ago

What is the difference between radial acceleration and tangential acceleration and how do you calculate both of these accelerati

sergey [27]

Answer:

Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion

a = (V2 - V1) / T

Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)

a = m V^2 / R = m ω^2 R where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not in direction

8 0

1 year ago

A swimmer jumps from a waterfall thats is 8.5 m high from the lake below. How fast must he run horizontally to dive 2.5m away fr

Mashutka [201]

He must run at 6m/s

6 0

3 years ago