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frosja888 [35]
3 years ago
8

Each cell in the battery has a potential difference of 1,5 V. calculate. the reading on the ammeter ,A​

Physics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

0.30A

Explanation:

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Write a list of at least five scientific methods.
kolbaska11 [484]
1- You should always have a question for your experiment.
2- You need to conduct background research. It helps to write down your sources so you can cite your references. 
3- Propose a hypothesis (educated guess on what you believe the outcome of the experiment will be)
4- Design and perform an experiment to test your hypothesis (include independent and dependent variable)
5- Record observations and analyze what the data means.
6- Conclude whether you need to accept or reject your hypothesis, which accepting means your hypothesis was right and rejected is if it was wrong. 
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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 
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3 years ago
What does the second law of throdynamics state?
Anna71 [15]

Explanation:

what does the second law of throdynamics state?

A. The amount of molecular disorder in the universe will always increase

B. The amount of molecular disorder in the universe will always decrease

C. The amount of molecular disorder in the universe will stay constant

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<span>The forth I think, conserving water is useless if it is just going to be contaminated in unsanitary conditons
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two vectors have a magnitude of 2.5km and 6.5 km . Predict the maximum and minimum magnitudes of their resultant vector
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The maximum magnitude of their resultant vector is when the two vectors are parallel and in the same direction, so they lie on the same axis. In this case, the magnitude of their resultant vector is simply the sum of the two magnitudes:

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The minimum magnitude of their resultant vector is when the two vectors are parallel but in opposite direction. In this case, the magnitude of their resultant vectors is just the difference between the two magnitudes:

R=6.5 km-2.5 km=4.0 km

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3 years ago
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