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2 years ago

A 60KG WOMAN IS ON A LADDER 2 M ABOVE THE GROUND. WHAT IS HER POTENTIAL ENERGY

Physics

1 answer:

dexar [7]2 years ago

7 0

Answer:

Explanation:

PE = mgh = 60(9.8)(2.0) = 1176 J

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A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

8 0

2 years ago

A horizontal line on a displacement-time graph means the object is

r-ruslan [8.4K]

Answer:

traveling further away

Explanation:

4 0

2 years ago

Select all that apply

borishaifa [10]

a,b,c is your answer light and sound are not considered matter and heat is energy created from matter and electricity is particles moving basically therefore electricity is matter hope this helps

5 0

2 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

7 0

3 years ago

In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002

vichka [17]

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

6 0

2 years ago