If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far along the length of the ladder can a 69.1-kg painter climb before the ladder begins to slip?

Answer:

d. interaction atmosphere and biosphere interaction

Explanation:

hydrosphere, lithosphere, and biosphere interaction.

Answer:

For the first one it would be screw and for the second one it is a ramp

Explanation:

Hope this helps

Answer:

A) 0.20Joules

Explanation:

The type mechanical energy acting on the body is potential energy since the body is covering a particular height.

Potential Energy = mass×acceleration due to gravity × height

Mass of the body = 0.1kg

acceleration due to gravity = 10m/s²

h is the height of the object when dropping = 1.0meters

Substituting this values in the formula to get the energy of the body on dropping, we have;

PE = 0.1×10×1.0

PE = 1.0Joules

On bouncing back to height of 0.8m, the potential energy becomes

PE = 0.1×10×0.8

PE = 0.8Joules

The mechanical energy lost by the ball as it bounces will approximately be the difference in its potential energy when dropping and when it bounces back i.e 1.0Joules - 0.8Joules = 0.20Joules

0.2 J

mgh (before) = (0.1)(9.8)(1) = 0.98

mgh (after) = (0.1)(9.8)(0.8) = 0.784

0.98-0.784=0.196

0.196 is approximately 0.2.  

Answer:

Explanation:

Given

acceleration of a particle is given by [tex]a(t)=pt^2-qt^3[/tex]

[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=pt^2-qt^3[/tex]

[tex]dv=(pt^2-qt^3) dt[/tex]

Integrating we get

[tex]\int dv=\int \left ( pt^2-qt^3\right )dt[/tex]

[tex]v=\frac{pt^3}{3}-\frac{qt^4}{4}+c_1[/tex]

at [tex]t=0\ v=0[/tex]

therefore [tex]c_1=0[/tex]

We know velocity is given by

[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]

[tex]vdt=dx[/tex]

integrating

[tex]\int dx=\int \left ( \frac{pt^3}{3}-\frac{qt^4}{4}+\right )dt[/tex]

[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}+c_2[/tex]

using conditions

at [tex]t=0\ x=0[/tex]

[tex]c_2=0[/tex]

[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}[/tex]                          

(a) The velocity of the particle as function of time is pt³/3 - qt⁴/4 + C.

(b) The position of the particle as function of time is pt⁴/12 - qt⁵/20 + Ct + C.

The velocity of the particle is the integral of the acceleration of the particle.

The velocity of the particle is calculated as follows;

v = ∫a(t)

v = ∫(pt² - qt³)dt

v = pt³/3 - qt⁴/4 + C

The position of the particle as function of time is the integral of velocity of the particle.

x = ∫v

x = ∫(pt³/3 - qt⁴/4 + C)dt

x = pt⁴/12 - qt⁵/20 + Ct + C

Learn more about velocity and position here: https://brainly.com/question/24445340

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

                   R = 2r

                      = [tex]2 \times 0.1 m[/tex]

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = [tex]\frac{k \times Q \times Q}{R}[/tex]

                    = [tex]\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}[/tex]

                    = 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, [tex]\frac{U}{2}[/tex].

Hence,   K.E = [tex]\frac{U}{2}[/tex]

                     = [tex]\frac{0.081}{2}[/tex]

                     = 0.0405 J

Now, we will calculate the speed of balls as follows.

                 V = [tex]\sqrt{\sqrt{\frac{2 \times K.E}{m}}[/tex]

                     = [tex]\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}[/tex]

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

Answer:

D.

Explanation:

It just makes sense. It is a simple answer

Answer: 0.5m/s

This value may vary depending on the initial velocity value.

Explanation:

The question is incomplete due to absence of the initial velocity (U) of the rock. Taking the initial velocity as any value say 10m/s at 30° above the horizontal.

Time it will take to reach maximum height (Tmax) will be Usin(theta)/g

Where U is the initial velocity = 10m/s theta = 30° g is the acceleration due to gravity = 10m/s²

Substituting the values in the formula we have;

Tmax = 10sin30°/10

Tmax = sin 30°

Tmax = 0.5second

Therefore, it will take for the rock 0.5s to reach the maximum height of its trajectory if the velocity is 10m/s.

The time varies though depending on the value of the initial velocity.

a)

i) Potential for r < a: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

ii) Potential for a < r < b:  [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]

iii) Potential for r > b: [tex]V(r)=0[/tex]

b) Potential difference between the two cylinders: [tex]V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

c) Electric field between the two cylinders: [tex]E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]

where

[tex]\lambda[/tex] is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

[tex]V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)[/tex]

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

[tex]E=0[/tex]

So the potential where the electric field is zero is constant:

[tex]V=const.[/tex]

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density [tex]+\lambda[/tex] and an equal negative charge density [tex]-\lambda[/tex]. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

[tex]\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)[/tex]

However, we know that the potential at b is zero, so

[tex]V(r)=V(b)=0[/tex]

ii) The electric field in the region a < r < b instead it is given only by the positive charge [tex]+\lambda[/tex] distributed over the surface of the inner cylinder of radius a, therefore it is

[tex]E=\frac{\lambda}{2\pi r \epsilon_0}[/tex]

And so the potential in this region is given by:

[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex] (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

And so, for r<a,

[tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

- Potential at the surface of the outer cylinder:

[tex]V(b)=0[/tex]

Therefore, the potential difference is simply equal to

[tex]V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]

The electric field is just the derivative of the electric potential:

[tex]E=-\frac{dV}{dr}[/tex]

so we can find it by integrating the expression for the electric potential. We find:

[tex]E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

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Answer:

A

Explanation:

Why is that they truly evaluate both of the theories in the same way.

Answer:

Explanation:

This question is quite tricky. Not all parameters were provided.

λ = 515nm(nm = namo meters), 515 nm becomes 515 x 10^-9 which in turn becomes 5.15 x 10^-7.

The velocity for the green light wasn't provided but the velocity of greenlight is a constant; velocity of green light(v) = 3.00 × 108 m/s.

frequency of the light =[tex]\frac{velocity of green light}{wavelength of light}\\[/tex]

f =[tex]\frac{3.00 X 108 m/s}{5.15 X 10^{-7} }[/tex]

f=20.971 Hertz

The frequency of light will be "20.971 Hz".

According to the question,

Wavelength,

or,

          [tex]= 5.15\times 10^{-7} \ m[/tex]

Velocity of green,

The frequency of light (f) will be:

= [tex]\frac{Velocity}{Wavelength}[/tex]

= [tex]\frac{3.00\times 108}{5.15\times 10^{-7}}[/tex]

= [tex]20.971 \ Hz[/tex]

Learn more:

https://brainly.com/question/14618870

Answer:

B. 10 days

Explanation:

Answer:

Option (3)

Explanation:

Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.

He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.

This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.

Thus, the correct answer is option (3).

Answer:

I think it is option 'c' in the quiz.

Explanation:

I took the test and got a 100.

Answer:

Explanation:

Given

Charge [tex]Q_1=+5.5\mu C[/tex]

[tex]Q_2=-17\ mu C[/tex]

When the two charges are touch then the net charge on the two sphere will be same after some time

[tex]Q=\frac{Q_1+Q_2}{2}[/tex]

[tex]Q=\frac{5.5-17}{2}=-5.75\ \mu C[/tex]

Now the force between this tow charges are when they are separated with [tex]d=60\ mm[/tex]

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

[tex]F=\frac{9\times 10^9\times (-5.75)^2}{(60\times 10^{-3})^2}[/tex]

[tex]F=82.65\ N[/tex]

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s[/tex]

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s