A car travels west at 40 km/h

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

3 years ago

The atomic number of a element if found by countingthe number of what in an atom?

S_A_V [24]

The Atomic number is found by counting the number of Protons found in the atom. Make brainliest if I helped best please :)

4 0

3 years ago

The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4

scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

v = v₀ + a t

y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing.

Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s

v = v₀ + a t

$a = \frac{v-v_o}{\Delta t}$

$a = \frac{2-0}{8-0}$

a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

$a = \frac{6-2}{ 14-10}$

a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

$d_{total } = d_1 +d_2 + d_3 +d_4$

We look for every displacement.

d₁ = v₀ + ½ a₁ Δt²

d₁ = 0 + ½ 0.25 8²

d₁ = 8 m

In the second section the velocity is constant

d₂ = v₂ Δt₂

d₂ = 2 (10-8)

d₂ = 4 m

The third section.

d₃ = v₀ + ½ a t²

d₃ = 2 + ½ 1 (14-10) ²

d₃ = 10 m

The distance of the fourth section.

we look for acceleration

a₄ = $\frac{v-v_o}{\Delta t}$

a₄ = $\frac{0-6}{18-14}$

a₄ = -1.5 m / s²

d₄ = 6 + ½ (-1.5) (1814) ²

d₄ = -6 m

The total displacement is;

$d_{total}$ = 8 + 4 + 10 -6

$d_{total}$ = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0

2 years ago

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

Elina [12.6K]

Before going to answer this question first we have to know the fundamental principle of magnetism.

A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.

Through this concept the question can be answered as explained below-

A-As per first option the side of magnet A is repelled by the south pole of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.

B-The side of magnet A is repelled by the north pole of magnet B. Hence the side of A must be north pole.It can't be a south pole.

C-The side of magnet A is attracted by the south pole of magnet B .Hence the side of magnet A must be north.Hence this is right

D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.

Hence the option C is right.

3 0

3 years ago

Read 2 more answers

A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the

Alecsey [184]

Answer:

Part a)

$KE = 77.95 J$