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3 years ago

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

Physics

1 answer:

Volgvan3 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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konstantin123 [22]

Answer:

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(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

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(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

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3 years ago

If i try to fail and succeed which one did I do

Novay_Z [31]

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3 years ago

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

Serhud [2]

Answer:

The value is $h = 13.2 \ m$

Explanation:

From the question we are told that

The speed of the rope with hook is $u = 24 .1 \ m/s$

The angle is $\theta = 65.0^o$

The speed at which it hits top of the wall is $v = 16.3 m/s$

Generally from kinematic equation we have that

$v_y^2 = u_y ^2 + * 2 (-g)* h$

Here h is the height of the wall so

$[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h$

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3 years ago

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

Kipish [7]

Answer:

F = 47.6 N

Explanation:

Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

$F = \frac{\Delta p}{\Delta t}$

So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:

$F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N$

⇒ Fsk = 47.6 N (normal to the wall)

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3 years ago

A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car

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Answer:

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Explanation:

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