A compound consisting of atoms of small atomic mass is more likely to require

A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio

Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0

4 years ago

5. An acrobat, starting from rest, swings freely on a trapeze of

34kurt

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

Em₀ = U = m g h

Final point. Lower part of the movement

$Em_f$ = K = ½ m v²

Energy is conserved.

Emo = $Em_f$

m g h = ½ m v²

v² = 2 gh

Let's use trigonometry to find the height, see attached.

h = L - L cos θ

h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

h = 3.7 (1- cos 48)

h = 1.22 m

this is the maximum height of the movement.

Let's calculate the velocity.

$v= \sqrt{2 \ 9.8 \ 1.22}$

v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

a) The maximum speed is v = 4.89 m / s

b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

5 0

3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)