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3 years ago

5

Use an energy method to derive the equation of motion of the system in Problem 1.112. First with the presence of the damper and

then without. (hint: your answer should match your equation of motion obtained from Newton’s second law).

1 answer:

aliya0001 [1]3 years ago

5 0

Answer and Explanation:

The answer is attached below

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What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

Alisiya [41]

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

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3 years ago

If BHALA AHMAD KHAN applied the 20N force is applied on an object moving with the velocity 30 m/s. calculate the power in KW.

Leokris [45]

Answer:0.6kw

Explanation:

Power=force×velocity

Power=20×30=600w

In kw it's going to be 600/1000=0.6kw

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3 years ago

A roller coaster is stationary at the top of a ramp.

Serjik [45]

Answer:

A

Explanation:

The roller coaster is stationary so the kinetic energy would be zero, but it is at the top of ramp so the potential energy would be high as its gravitational so it would have to be A

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2 years ago

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

so maximum possible angle of the inclined plane is

$\theta = tan^{-1}\mu$

3 0

3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

6 0

3 years ago