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3 years ago

Are the objects described here in static equilibrium, dynamic equilibrium, or not equilibrium at all? Explain.

2 answers:

4 0

Let us examine the given situations one at a time.

Case a. A 200-pound barbell is held over your head.
The barbell is in static equilibrium because it is not moving.
Answer: STATIC EQUILIBRIUM

Case b. A girder is being lifted at a constant speed by a crane.
The girder is moving, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case c: A jet plane has reached its cruising speed at an altitude.
The plane is moving at cruising speed, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case d: A box in the back of a truck doesn't slide as the truck stops.
The box does not slide because the frictional force between the box and the floor of the truck balances out the inertial force. The box is in static equilibrium.
Answer: STATIC EQUILIBRIUM

nordsb [41]3 years ago

4 0

Statement a and e are in static equilibrium, b and d are in dynamic equilibrium and c is not at equilibrium at all.

<h3>FURTHER EXPLANATION</h3>

When the net force acting on an object is zero, the object is said to be at equilibrium. The state of equilibrium can be classified into two: static and dynamic.

Static equilibrium is when the net force is zero resulting in the object being at rest or not moving.

Dynamic equilibrium is when the resultant force acting on an object is zero and the object is moving in a uniform motion (i.e. constant or unchanging speed).

A. a 200 pound barbell is held above your head

<em>Since the barbel is "held above your head" and is implied to be at rest since there is no change in its position during the time it is held, then this situation is an example of static equilibrium.</em>

B. A girder is being lifted at a constant speed by a crane

<em>This is dynamic equilibrium because the object is moving "at a constant speed".</em>

C. A girder is being lowered into place. It is slowing down.

<em>This is not equilibrium condition because the objects is neither at rest nor moving at a constant speed. It is "slowing down", so the speed is decreasing.</em>

D. A jet plane has reached its cruising speed and altitude.

<em>The "cruising speed" is the speed that is maintained while the jet plane is traveling because this is considered to be the most efficient speed. Since it is maintained, the jet plane is said to be in constant motion. Therefore, this is dynamic equilibrium.</em>

E. A box in the back of a truck doesn't slide as the truck stops.

<em>This is static equilibrium because the box is stationary or not moving.</em>

<em />

<h3>LEARN MORE</h3>

Balanced Forces brainly.com/question/1675020
Resultant Force brainly.com/question/7041906
Friction brainly.com/question/3401004

Keywords: equilibrium, static equilibrium, dynamic equilibrium

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Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

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a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

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