What is most likely to result immediately after a rainforest in Brazil is clear-cut?secondary succession

A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

So

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

So

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

3 0

2 years ago

If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth

Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

F_b = Y * V_timber

F_b = 9810*0.15*0.15*x

F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

(M)_p = 0

W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

670*0.5*3.6 - x^2 * 0.5*226.7 = 0

x^2 = 1206 / 113.35

x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

sin(Q) = y / x

sin(Q) = 2.1 / 3.26

Q = sin^-1 (0.6441718)

Q = 40.1 degrees

5 0

3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

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

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC

nignag [31]

Answer:

ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

ΔL = α L₀ ΔT

ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

ΔL = 3.82 10⁻⁴ m

5 0

3 years ago

A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (

Margarita [4]

$\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

4 0

2 years ago