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larisa [96]
3 years ago
14

A fan is designed to last for a certain time before it will have to be replaced (planned obsolescence). The fan only has one spe

ed (at a maximum of 700 rpm), and it reaches the speed in 5.0 s (starting from rest). It takes the fan 7.0 s for the blade to stop once it is turned off. The manufacturer specifies that the fan will operate up to 1 billion rotations. Andre lives in a hot climate, works outside of the home from approximately 8:00 am to 5:00 pm, Monday through Friday, does not own an air conditioner, and can't sleep with the fan running. Estimate how many hot days N hot Andre will be able to use the fan, rounded to the nearest day.
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

Answer:

Explanation:

<u>Total Minutes</u>

The fan can achieve 1,000,000,000 rotations in its lifetime and does 700 rotations per minute

Therefore, the fan will work for \frac{1,000,000,000}{700} = 1,428,571.45 minutes in its lifetime.

<u>Working Time</u>

Andre works for 9 hours, assuming he sleeps for 8 hours, The fan is on for 7 hours on weekdays (25 - 9 - 8), a total of 35 hours and

16 hours on the weekend (24 - 8), a total of 32 hours

The fan is on for 67 hours a week or 4020 minutes.

<u>Downtime</u>

Everyday the fan is used, it takes 5.0 seconds to get to its full speed and 7.0 seconds for the blade to stop, an additional 2 seconds per day. This adds 14 seconds per week.

In a week, the fan works for 4020 minutes + (\frac{14}{60})

=4020.23minutes

In its lifetime, the fan will work for \frac{1,428,571.45 }{4020.23} weeks.

=355.35 weeks

OR 355.35 X 7 Days

2487 hot days

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In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

tan \theta =\frac{A}{B}=\frac{10}{10}=1

\theta=tan^{-1} (1)=45^{\circ}

and the direction will be north-east.

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What does a animal and a plant cell have in common.
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Answer:

Structures that are common to plant and animal cells are the cell membrane, nucleus, mitochondria, and vacuoles. Structures that are specific to plants are the cell wall and chloroplasts.

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In general, how do you find the average velocity of any object falling in a vacuum?
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3 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
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A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

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The solutions and homogeneous mixtures in the given list are:

A. Lead solder, an alloy of tin and lead.

D. Window cleaner, a mixture of ammonia and coloring dissolved in water.

E. Gasoline, a mixture of organic liquids with a fixed composition throughout.

<h3>What is a solution?</h3>

A solution can be defined as a special type of homogeneous mixture that comprises a solute and a solvent.

A homogeneous mixture can be defined as any solid, liquid, or gaseous mixture which has an identical (uniform) composition and properties throughout any given sample of the mixture.

In Science (Physics), all solutions are considered to be a homogeneous mixture because their constituents are uniformly (evenly) distributed.

In conclusion, the solutions and homogeneous mixtures in the given list are:

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