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3 years ago

9

An empty oﬃce chair is at rest on a ﬂoor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

e exerted by the ﬂoor; and. 3. A net downward force exerted by the air. Which force(s) act on the oﬃce chair?. 1. None of the forces act on the chair since. it is at rest. 2. All three forces. 3. 1 only. 4. 1 and 2 only. 5. 2 and 3 only.

2 answers:

LenaWriter [7]3 years ago

6 0

The forces acting on the chair are:

4. 1 and 2 only.

Julli [10]3 years ago

6 0

The forces that are acting on the chair are the downward force due to gravity and the upward force exerted by the floor. The correct option among all the options that are given in the question is the fourth option or option "4". I hope that this answer has come to to your help.

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Which of the following is a device designed to open an overloaded circuit and prevent overheating?

spin [16.1K]

Circuit breaker is a device that is designed to open an overloaded circuit and prevent overheating. The correct option in regards to the given question is option "a". The main purpose of the circuit breaker is to disconnect the defective switch from the circuit as soon as any problem arises. This helps in preventing any kind of major accident. Overheating can result in a big accident and the circuit breaker is the device that senses the overheating and disconnects the switch before any major incident. It is an automated system that acts as a protective devise.

4 0

3 years ago

Scenario 2: Use the following information to answer questions 3 and 4:

NemiM [27]

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

$\dfrac{MET*weight (kg)*3.5}{200}$

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = $\dfrac{4.3*75 (kg)*3.5}{200}$

= 5.644

Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.

4.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

$V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7$

where ;

$V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = 0.0012$

∴ $0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min$

Converting to watts;

Since; 6.118kg-m/min is = 1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

≅ 47 watts

3 0

4 years ago

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

$\tau=F\times r$

$\tau=82.0\times0.150$

$\tau=12.3\ N-m$

Now, The angular acceleration

$\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}$

$\dfrac{d\omega}{dt}=20.73\ rad/s^2$

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

$\tau=I\times\dfrac{d\omega}{dt}$

$I=\dfrac{I}{\dfrac{d\omega}{dt}}$

$I=\dfrac{12.3}{20.73}$

$I= 0.593\ kg-m^2$

Hence, The moment of inertia of the wheel is 0.593 kg-m².

5 0

3 years ago

Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Part

Iteru [2.4K]

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

$F_g = \frac{Gm_1m_2}{r^2}$

Where,

G = Gravitational Universal Force

$m_i =$ Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

$m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m$

Applying the previous equation at X-Axis,

$F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N$

Applying the previous equation at Y-Axis,

$F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N$

Therefore the angle can be calculated as,

$tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°$

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

5 0

4 years ago

1. Describe the vibration caused by primary waves.

mrs_skeptik [129]

Energy released during an earthquake travels in the form of waves<span> around the Earth. Two types of seismic </span>wave<span> exist, </span>P<span>- and </span>S-waves<span>. They are different in the way that they travel through the Earth. ... They are transverse </span>waves<span> which mean the </span>vibrations<span> are at right angles to the direction of travel.</span>

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4 years ago