Compare and contrast what happens to an object's motion when balanced or

The weight of the block in the drawing is 97.0 N. The coefficient of static friction between the block and the vertical wall is

katrin [286]

Hello,

I hope you're having a great day!

Here's what I got to question A and B

a) Fs(max)=(0.560)(88.9N)=49.8N Fy=88.9N-49.8N=39.1N Fx=(39.1N)/(cos(40))=51.0N sqrt{39.1^2+51.0^2}=F

F=64.3N

b) 88.9N +49.8 N=138.7N=Fy Fx=(138.7N)/(cos(40))=181.06N sqrt{138.7^2+181.06^2}=F

F=228N

4 0

2 years ago

1.do materials change when light is reflected off to them

hodyreva [135]

Number two) C. Maybe

7 0

2 years ago

Read 2 more answers

Consider a string of total length L, made up of three segments of equal length. The mass per unit length of the first segment is

zzz [600]

Answer:

Explanation:

Length of each segment is $\frac{L}{3}$

Speed of wave in first segment is $v_1=\sqrt{\frac{T_s}{\mu}}$

Speed of wave in second segment is $v_2=\sqrt{\frac{T_s}{2\mu}}$

Speed of wave in third segment is $v_3=\sqrt{\frac{T_s}{\frac{\mu}{4}}}=\sqrt{\frac{4T_s}{\mu}}$

Now time for the transverse wave to propagate is

$t=t_1 + t_2 + t_3\\=\frac{(\frac{L}{3})}{v_1}+\frac{(\frac{L}{3})}{v_2} + \frac{(\frac{L}{3})}{v_3}\\\\=(\frac{L}{3})(\frac{1}{\sqrt{\frac{T_s}{\mu}}} + \frac{1}{\sqrt{\frac{T_s}{2\mu}}} + \frac{1}{\sqrt{\frac{4T_s}{\mu}}})$

simplifying we get

$t=(\frac{3+2\sqrt{2}}{6})L\sqrt{\frac{\mu}{T_s}}$

3 0

2 years ago

If a spring is stretched to twice the lengtj of its equilibrium position, by what factor does the energy stored in the spring ch

mylen [45]

Answer:

the energy increases 4 times

Explanation:

A spring has an elastic potential energy that is given by the expression

K_e = ½ K (x-x₀)²

where x is the distance from the equilibrium point and k is the return constant

if the spring is stretched at x-x₀ = 2x₀, the energy value

K_e = ½ k (2x₀)²

K_e = ½ k 4 x₀²

K_e = 4 (½ k x₀²)

$\frac{K_e}{ \frac{1}{2} k x_o^2}$ = 4

therefore the energy increases 4 times

6 0

2 years ago

4 points

zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

$I = k\frac{1}{r^2}$

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

$I_2 = k\frac{1}{(2r)^2}$

Dividing both the equation we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4} = 25lx$

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

$I_2 = k\frac{1}{(3r)^2}$

Dividing equation 1 and 3 we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9} = 11.11lx$

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0

2 years ago