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A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

3 years ago

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$

Explanation:

We have given mass of the air particle $m=5\times 10^{-16}kg$

Gas constant R = 8.314 J/mol-K

Temperature is given T = $27^{\circ}C=273+27=300K$

We have to find the root mean square speed of the particle

Which is given by $v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec$

So rms speed of the particle will be $38.68\times 10^8m/sec$

5 0

3 years ago

. The spot on Jupiter is a _______? Volcano Ocean Storm Unknown

MaRussiya [10]

Storm!! Hope I could help

8 0

3 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is compressed at steady state to 12 bar, 80°C. The vol

3241004551 [841]

Answer:

The magnitude of the heat transfer rate from the compressor is 87.05 kW

Explanation:

Initial pressure of refrigerant = 4 bar

Final pressure of refrigerant = 12 bar

From steam table,

Internal energy at 4 bar (U1) = 2554 kJ/kg

Internal energy at 12 bar (U2) = 2588 kJ/kg

Change in internal energy (∆U) = U2 - U1 = 2588 - 2554 = 34 kJ

Work input (W) = 120 kJ/kg

Quantity of heat transfer (Q) = ∆U + W = 34 + 120 = 154 kJ/kg

Volumetric flow rate of refrigerant = 8 m^3/min = 8/60 = 0.133 m^3/s

Density of refrigerant = 4.25 kg/m^3

Mass flow rate = density × volumetric flow rate = 4.25 kg/m^3 × 0.133 m^3/s = 0.56525 kg/s

Q = 154 kJ/kg × 0.56525 kg/s = 87.05 kJ/s = 87.05 kW

6 0

3 years ago

Explain how you would apply any five pricing techniques to attract customers

Anna11 [10]

Answer:

Explanation:

1. DECOY PRICING

This occurs when customers make a purchase they must often choose between products with different prices and attributes.

This method of pricing is meant to influence the consumer's purchasing decision and maximise the sales of one particular product. The seller will offer at least three products; two of the products will have a similar or equal price. The two products with the similar prices should be the most expensive ones, and one of the two should be less attractive than the other.

2. LOSS LEADER

This is when a product is sold at a low price (often without profit) in order to stimulate other profitable sales or to attract new customers.

The main is that it will help the business to expand their market share as a whole. It's common practice when first entering a market as it introduces new customers to a service or product in the hope of building a customer base and securing future

3.ODD PRICING

This is a method of psychological pricing a product. Prices ending in 9, 95, 97, 99 are sometimes called “charm prices” and in this type of pricing, the seller fixes a price where the last digits are odd numbers. This is intended to give the buyer no room for manœuvering or for bargaining as the price appears to be less - a product priced at £9.99 will seems much cheaper than one priced at £10.00

4. PRICE DISCRIMINATION

The purpose of price discrimination is to capture the market's consumer surplus and generate the most revenue possible for a product. Identical goods or services are sold at different prices from the same provider to different segments of the market. Industries that commonly use price discrimination include the travel industry, pharmaceuticals and textbook publishers.

5. PRODUCT BUNDLE PRICING

Using this method, sellers will combine several products in the same package. It also serves to move old stock. Blu-ray and videogames are often sold using the bundle approach once they reach the end of their product life cycle. This technique is used at auctions where one attractive item may be included in a lot with a box of less interesting things. Buyers must bid for the entire lot. It’s a good way of moving slow selling products, and in a way is another form of promotional pricing.

4 0

2 years ago