? energy that results from the position or shape of an object , Potential energy related to an objects ?

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

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the

ValentinkaMS [17]

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

5 0

3 years ago

How much louder (more intense) is a 48-dB sound than a 10-dB sound?

worty [1.4K]

Answer:

10jwhsjsjjsiwiiqiqiw

8 0

3 years ago

Read 2 more answers

Anybody help me? my H.W. Q 14-17 thank you, please show work

Kipish [7]

14. Reactants:

Magnesium Chloride: $Mg Cl_2$ , since magnesium (Mg) has 2 valence electrons while chlorine (Cl) has 1 vacancy, so they are in ratio 1:2

Sodium oxide: $Na_2 O$ , since sodium (Na) has 1 valence electron while oxygen (O) has 2 vacancies, so they are in ratio 2:1

15. Products:

Magnesium oxide: $Mg O$ , since magnesium (Mg) has 2 valence electrons and oxygen (O) has 2 vacancies, so they are in ratio 1:1

Sodium chloride: $NaCl$ , since sodium (Na) has 1 valence electron and chlorine (Cl) has 1 vacancy, so they are in ratio 1:1

16: Balanced reaction:

$Mg Cl_2 + Na_2 O \rightarrow Mg O+2NaCl$

we see that on the reactant side we have: 1 atom of Mg, 2 atoms of Cl, 2 atoms of Na and 1 atom of O. By putting a 2 in front of NaCl, we have exactly the same on the product side: 1 atom of Mg, 2 atoms of Cl, 2 atoms of Na and 1 atom of O.

17. It is a double-replacement reaction

A double-replacement reaction occurs when two ionic compounds exchange their ions. In this specific case, we see that the Chlorine (initially bond with the magnesium) is then bond with the sodium, while the oxygen (initially bond with the sodium) after the reaction is bond with the magnesium.

4 0

3 years ago

Two charged objects have a repulsive force of .080 N. If the charge of both of the objects is doubled, then what is the new forc

irina1246 [14]

heyy have a nice day yixxixigx

7 0

2 years ago