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2 years ago

A 205 kg log is pulled up a ramp by means of a rope that

Physics

1 answer:

Volgvan2 years ago

7 0

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination $=30^{\circ}$

coefficient of kinetic friction between log and ramp is ( $\mu _k$ )=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

$T-mgsin\theta -f_r=ma$

Where $mgsin\theta$ =Sin component of weight

$f_r=friction\ Force=\mu _KN$ (Where N is Normal reaction)

$T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma$

$T=m\left ( gsin\theta +\mu _kcos\theta +a\right )$

sin30=0.5

cos30=0.866

$T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )$

$T=205\times 13.35=2.737 kN$

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2 years ago

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3 years ago

In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the f

zhannawk [14.2K]

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

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2 years ago

A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

$I = 0.106 \ A$

Explanation:

From the question we are told that

The current is $I = 0.106 \ A$

The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

The separation between the plate is $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

$\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value

$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

$I = 8.85*10^{-12} * 1.1977 *10^{10}$

=> $I = 0.106 \ A$

3 0

2 years ago

Find mass when kinetic energy is 1.6 J and velocity is 0.2 m/s

klio [65]

Answer:

4kg

Explanation:

Given parameters:

Kinetic energy = 1.6J

Velocity = 0.2m/s

Unknown:

Mass of the body = ?

Solution:

The kinetic energy of a body is the energy due to the motion of the body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

1.6 = $\frac{1}{2}$ x 0.2 x v²

1.6 = 0.1v²

v² = 16

v = 4kg

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2 years ago