A person is pushing 108 kg box with a 325 n force if the friction force is 55 n what is the acceleration of the box

Lyrx [107]

Answer:

<em>1</em><em>. </em><em>A body is said to be at rest if its position does not change with respect to its surroundings.</em>

8 0

4 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

In an experiment to determine the s.h.c. of lead, a 0.80 kg block of lead is heated using a

gladu [14]

Answer:2.47

Explanation: did the math

8 0

2 years ago

Find the force required to do 25 joule work when the force causes a displacement of 0.5 m

Eduardwww [97]

Answer:

Explanation:

The force required can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{25}{0.5} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

The y-position of a damped oscillator as a function of time is shown in the figure.

NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0

2 years ago