PLEASE HELP ASAP!!!!!
Carrie pulls a 41.5 kg bin across the floor of her garage using a rope that is attached to the bin. She pulls with a force of 138 N, at an angle of 28.0° above horizontal. The coefficient of kinetic friction between the floor and the box is 0.290 . What is the acceleration of the box?

Answer:

12 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bag = 25 kg

Potential energy (PE) = 2940 J

Height (h) =?

Objects carried to a particular height will always experience an acceleration due to gravity of 9.8 m/s².

With the above in mind, we can obtain the height to which the load is lifted to as shown below:

Mass (m) of bag = 25 kg

Potential energy (PE) = 2940 J

Acceleration due to gravity (g) = 9.8 m/s².

Height (h) =?

PE = mgh

2940 = 25 × 9.8 × h

2940 = 245 × h

Divide both side by 245

h = 2940 / 245

h = 12 m

Therefore, the worker lifts the load to a height of 12 m.

Answer:

The torque necessary to bring the wheel to rest in 40 seconds is 10.4 N·m

Explanation:

The question is with regards to rotational motion

The rotary motion parameters are;

The mass of the wheel = 50 kg

The radius of the wheel = 0.4 m

The rate of rotation of the wheel = 480 rpm

The time in which the wheel is to be brought to rest = 40 s

The rotational rate of the wheel in rotation per second is given as follows;

480 r.p.m = 480 r.p.m × 1 minute/(60 seconds) = 8 revolution/second

1 revolution = 2·π radians

Therefore, we have the angular velocity, ω, given as follows;

ω = 2·π × 8 revolutions/second ≈ 50.3 rad/s

The angular acceleration, α, is given as follows;

[tex]\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{\omega _2 - \omega_1}{t_2 - t_1}[/tex]

Whereby the wheel is brought to rest from its initially constant rotational motion in 40 seconds, we have;

ω₁ ≈ 50.3 rad/s, ω₂ = 0 rad/s, and t₂ - t₁ = 40 seconds

Plugging in the values for the variables of the equation for the angular acceleration, "α", we get;

[tex]\alpha = \dfrac{0 - 50.3 \ rad/s}{40 \ s} \approx 1.3 \ rad/s^2[/tex]

The torque on the wheel, τ, is given as follows;

τ = m·r²·α

Where;

m = The mass of the object = 50 kg

r = The radius of the wheel = 0.4 m

α = The acceleration of the wheel ≈ 1.3 rad/s²

Therefore;

τ = 50 kg × (0.4 m)² × 1.3 rad/s² ≈ 10.4 N·m

The torque necessary to bring the wheel to rest in 40 seconds = τ ≈ 10.4 N·m.

Answer:

-10.048 N m

Explanation:

Answer:

elements

not really an explanation

Answer:

3rd law beacuse there a flies